SOLUTIONS TO LIMITS USING L'HOPITAL'S RULE

SOLUTION 1 :

$${ \lim_{x \to 1} \ {x^2-1 \over x^2+3x-4} }$$ = $${ \lq\lq \ 1-1 \ '' \over 1+3-4 }$$ = $${ \lq\lq \ 0 \ '' \over 0 }$$

(Apply Theorem 1 for l'Hopital's Rule. Differentiate top and bottom separately.)

= $${ \lim_{x \to 1} \ {2x \over 2x+3} }$$

truein

= $${ {2(1) \over 2(1)+3} }$$

truein

= $${ 2 \over 5 }$$ . truein truein SOLUTION 2: $${ \lim_{x \to 4} \ {x-4 \over \sqrt{x}-2} }$$ = $${ \lq\lq \ 4-4 \ '' \over \sqrt{4} - 2 }$$ = $${ \lq\lq \ 0 \ '' \over 0 }$$ truein (Apply Theorem 1 for l'Hopital's Rule. Differentiate top and bottom separately.) truein

= $${ \lim_{x \to 4} \ {1 \over { 1 \over 2 \sqrt{x} }} }$$

truein

= $${ \lim_{x \to 4} \ { 2 \sqrt{x} } }$$

truein

= $${2 \sqrt{4} }$$

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= $4$ .

truein truein SOLUTION 3: $${ \lim_{x \to 0} \ {\sin x \over x} }$$ = $${ \lq\lq \ \sin 0 \ '' \over 0 }$$ = $${ \lq\lq \ 0 \ '' \over 0 }$$ truein (Apply Theorem 1 for l'Hopital's Rule. Differentiate top and bottom separately.) truein

= $${ \lim_{x \to 0} \ { \cos x \over 1 } }$$

truein

= $${ \cos 0 }$$

truein

= $${1 }$$ .

truein truein SOLUTION 4: $${ \lim_{x \to 0} \ {3^x-2^x \over x^2-x} }$$ = $${ \lq\lq \ 3^{0} - 2^{0} \ '' \over 0-0 }$$ = $${ \lq\lq \ 1 - 1 \ '' \over 0 }$$ = $${ \lq\lq \ 0 \ '' \over 0 }$$ truein (Apply Theorem 1 for l'Hopital's Rule. Differentiate top and bottom separately. Recall that $D \{a^x\} = a^x \ln a$.) truein

= $${ \lim_{x \to 0} \ { 3^x \ln 3 - 2^x \ln 2 \over 2x-1 } }$$

truein

= $${ { 3^{0} \ln 3 - 2^{0} \ln 2 \over 0-1 } }$$

truein

= $${ { (1) \ln 3 - (1) \ln 2 \over -1 } }$$

truein

= $${ \ln 2 - \ln 3 }$$

truein (Recall that $\ln A - \ln B = \ln(A/B)$.) truein truein

= $${ \ln(2/3) }$$ . truein truein SOLUTION 5: $${ \lim_{x \to 3} \ {{1/x} - {1/3} \over x^2-9} }$$ = $${ \lq\lq \ {1/3} - {1/3} \ '' \over 9-9 }$$ = $${ \lq\lq \ 0 \ '' \over 0 }$$ truein (Apply THEOREM 1 for l'Hopital's Rule. Differentiate top and bottom separately.) truein

= $${ \lim_{x \to 3} \ {{-1/x^2} \over 2x } }$$

truein

= $${ \lim_{x \to 3} \ { (-1/x^2)(1/2x) } }$$

truein

= $${ \lim_{x \to 3} \ { -1/2x^3 } }$$

truein

= $${ { -1/2(27) } }$$

truein

= $${ -1/54 }$$ .

truein truein

SOLUTION 6: $${ \lim_{x \to 0} \ {x \tan x \over \sin 3x } }$$ = $${ \lq\lq \ 0 \cdot \tan 0 \ '' \over \sin 0 }$$ = $${ \lq\lq \ 0 \cdot 0 \ '' \over 0 }$$ = $${ \lq\lq \ 0 \ '' \over 0 }$$ truein (Apply Theorem 1 for l'Hopital's Rule. Differentiate the top using the product rule and the bottom using the chain rule.) truein

= $${ \lim_{x \to 0} \ {x \sec^2 x+ (1) \tan x \over 3 \cos 3x } }$$

truein

= $${ { 0 \cdot \sec^2 0 + \tan 0 \over 3 \cos 0 } }$$

truein

= $${ { (0)(1)^2 + 0 \over 3(1) } }$$

truein

= $${ 0 \over 3 }$$

truein

= $${ 0}$$ .

truein truein SOLUTION 7: $${ \lim_{x \to 0} \ {\arcsin 4x \over \arctan 5x } }$$ = $${ \lq\lq \ \arcsin 0 \ '' \over \arctan 0 }$$ = $${ \lq\lq \ 0 \ '' \over 0 }$$ truein (Apply Theorem 1 for l'Hopital's Rule. Differentiate the top using the chain rule and the bottom using the chain rule. Recall that $D \{\arcsin f(x) \} = \displaystyle{1 \over \sqrt{ 1-(f(x))^2 } } \cdot f'(x)$ and $D \{\arctan f(x) \} = \displaystyle{1 \over 1+(f(x))^2 } \cdot f'(x)$ .) truein

= $${ \lim_{x \to 0} \ { { 1 \over \sqrt{1-(4x)^2} } (4) \over { 1 \over 1+(5x)^2 } (5) } }$$

truein

= $${ \lim_{x \to 0} \ { 4 \over \sqrt{1-16x^2} } \cdot { { 1+25x^2 } \over 5 } }$$

truein

= $${ { 4 \over \sqrt{1-0} } \cdot { { 1+0 } \over 5 } }$$

truein

= $${ { 4 \over 5 } }$$ .

truein truein SOLUTION 8: $${ \lim_{x \to 0} \ {\sin x^2 \over x\tan x} }$$ = $${ \lq\lq \ \sin 0 \ '' \over 0 \cdot \tan 0 }$$ = $${ \lq\lq \ 0 \ '' \over 0 \cdot 0 }$$ = $${ \lq\lq \ 0 \ '' \over 0 }$$ truein (Apply Theorem 1 for l'Hopital's Rule. Differentiate the top using the chain rule and the bottom using the product rule.) truein

= $${ \lim_{x \to 0} \ {2x \cos x^2 \over x \sec^2 x + \tan x} }$$

truein

= $${ {0 \cdot \cos 0 \over 0 \cdot \sec^2 0 + \tan 0 } }$$ = $${ {(0)(1) \over (0)(1)^2 + (0) } }$$ = $${ \lq\lq \ 0 \ '' \over 0 }$$

truein (Apply Theorem 1 for l'Hopital's Rule again. Differentiate the top using the product rule and the chain rule and the bottom using the product rule and the chain rule.) truein

= $${ \lim_{x \to 0} \ {2x(2x)(- \sin x^2 )+ 2 \cos x^2 \over ( x \cdot 2\sec x \cdot \sec x \tan x + \sec^2 x ) + \sec^2 x } }$$

truein

= $${ \lim_{x \to 0} \ {-4x^2 \sin x^2 + 2 \cos x^2 \over 2x \sec^2 x \tan x + 2 \sec^2 x } }$$

truein

= $${ { 0 \cdot \sin 0+ 2 \cos 0 \over 0 \cdot \sec^2 0 \cdot \tan 0 + 2 \sec^2 0 } }$$

truein

= $${{(0)(0)+ 2(1) \over (0)(1)^2(0) + 2(1)^2 } }$$

truein

= $${ 2 \over 2 }$$

truein

= $${1 }$$ .

truein truein SOLUTION 9: $${ \lim_{x \to 0} \ { x^2 e^x \over \tan^2 x} }$$ = $${ \lq\lq \ 0 \cdot e^{0} \ '' \over \tan^2 0 }$$ = $${ \lq\lq \ 0 \cdot 1 \ '' \over (0)^2 }$$ = $${ \lq\lq \ 0 \ '' \over 0 }$$ truein (Apply Theorem 1 for l'Hopital's Rule. Differentiate the top using the product rule and the bottom using the chain rule.) truein

= $${ \lim_{x \to 0} \ {x^2 e^x + 2xe^x \over 2 \tan x \cdot \sec^2x} }$$

truein

= $${ {(0)^2 e^{0} + 2(0) e^{0} \over 2 \tan 0 \cdot \sec^2 0 } }$$ = $${ {(0)(1)+(0)(1) \over (0)(1)^2 } }$$ = $${ \lq\lq \ 0 \ '' \over 0 }$$

truein (Apply Theorem 1 for l'Hopital's Rule again. Differentiate the top using the product rule and the bottom using the product rule and the chain rule.) truein

= $${ \lim_{x \to 0} \ {(x^2e^x+2xe^x)+(2xe^x+2e^x) \over 2 \tan x \cdot 2 \sec x \cdot \sec x \tan x + 2 \sec^2 x \cdot \sec^2 x } }$$

truein

= $${ \lim_{x \to 0} \ {x^2e^x+4xe^x+2e^x \over 4 \tan^2 x \cdot \sec^2 x + 2 \sec^4 x } }$$

truein

= $${ {(0)^2 e^{0} + 4(0) e^{0}+2e^{0} \over 4 \tan^2 0 \cdot \sec^2 0 + 2\sec^4 0 } }$$

truein

= $${ {(0)(1)+(0)(1)+2(1) \over 4(0)^2 (1)^2+2(1)^4 } }$$

truein

= $${ 2 \over 2 }$$

truein

= $1$ .

truein truein SOLUTION 10: $${ \lim_{x \to 0} \ { e^{-1/x^2} \over x^2 } }$$ = $${ \lq\lq \ e^{-1/0^2} \ '' \over (0)^2 }$$

truein

= $${ \lq\lq \ e^{-\infty} \ '' \over 0 }$$ = $${ \lq\lq \ { 1 / e^{\infty} } \ '' \over 0 }$$ = $${ \lq\lq \ { 1 / \infty } \ '' \over 0 }$$ = $${ \lq\lq \ 0 \ '' \over 0 }$$

truein (Apply Theorem 1 for l'Hopital's Rule. Differentiate the top using the chain rule and the bottom. Recall that $D \{ e^{f(x)} \} = e^{f(x)} \cdot f'(x)$ .) truein

= $${ \lim_{x \to 0} \ { e^{-1/x^2} \cdot 2/x^3 \over 2x } }$$

truein

= $${ \lim_{x \to 0} \ { 2e^{-1/x^2} \over x^3} \cdot { 1 \over 2x } }$$

truein

= $${ \lim_{x \to 0} \ { e^{-1/x^2} \over x^4 } }$$ = $${ \lq\lq \ 0 \ '' \over 0 }$$ .

truein (Applying Theorem 1 for l'Hopital's Rule again leads to $${ \lim_{x \to 0} \ { e^{-1/x^2} \over 2x^6 } }$$. Stop and think. This is getting nowhere. Go back to the beginning and algebraically rewrite the problem by ``flipping" both numerator and denominator.)

truein $${ \lim_{x \to 0} \ { e^{-1/x^2} \over x^2 } }$$ = $${ \lim_{x \to 0} \ { 1/x^2 \over e^{1/x^2} } }$$

truein = $${ { \lq\lq \ 1/0^2 \ '' \over e^{1/0^2} } }$$ = $${ { \lq\lq \ \infty \ '' \over e^{\infty} } }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule.) truein

= $${ \lim_{x \to 0} \ { -2/x^3 \over e^{1/x^2} \cdot -2/x^3 } }$$

truein

= $${ \lim_{x \to 0} \ { { 1 \over e^{1/x^2} } } }$$

truein = $${ \lq\lq \ 1 \ '' \over \infty }$$

truein = $0$ .

truein truein SOLUTION 11: $${ \lim_{x \to \infty} \ { e^{3x} \over 5x+200} }$$ = $${ \lq\lq \ e^{3 \cdot \infty} \ '' \over 5 \cdot \infty + 200 }$$ = $${ \lq\lq \ e^{ \infty} \ '' \over \infty }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule. Differentiate the top using the chain rule and the bottom.) truein

= $${ \lim_{x \to \infty} \ { 3e^{3x} \over 5 } }$$

truein

= $${ { 3e^{\infty} \over 5 } }$$

truein

= $${ { {\infty} \over 5 } }$$

truein

= $${ \infty }$$ (So limit does not exist.).

truein truein SOLUTION 12: $${ \lim_{x \to \infty} \ { 3 + \ln x \over x^2+7 } }$$ = $${ \lq\lq \ 3+\ln(\infty) \ '' \over (\infty)^2+7 }$$ = $${ \lq\lq \ 3+ \infty \ '' \over \infty + 7 }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule. Differentiate the top and the bottom.) truein

= $${ \lim_{x \to \infty} \ { 1/x \over 2x } }$$

truein

= $${ \lim_{x \to \infty} \ {1 \over x} \cdot {1 \over 2x} }$$

truein

= $${ \lim_{x \to \infty} \ {1 \over 2x^2 } }$$

truein

= $${ { \lq\lq \ 1 \ '' \over {\infty} } }$$

truein

= $${ 0}$$ .

truein truein SOLUTION 13: $${ \lim_{x \to \infty} \ {x^2+3x-10 \over 7x^2-5x+4} }$$ = $${ \lim_{x \to \infty} \ {x(x+3)-10 \over x(7x-5)+4} }$$

truein

= $${ \lq\lq \ (\infty)(\infty)-10 \ '' \over (\infty)(\infty)+4 }$$ = $${ \lq\lq \ \infty-10 \ '' \over \infty+4 }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule. Differentiate the top and the bottom.) truein

= $${ \lim_{x \to \infty} \ { 2x+3 \over 14x-5 } }$$

truein

= $${ \lq\lq \ 2(\infty)+3 \ '' \over 14(\infty)-5 }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule again.) truein

truein

= $${ \lim_{x \to \infty} \ {2 \over 14} }$$

truein

= $${ 1 \over 7}$$ .

truein truein SOLUTION 14: $${ \lim_{x \to \infty} \ { (\ln x)^2 \over e^{2x} } }$$ = $${ \lq\lq \ (\ln(\infty))^2 \ '' \over e^{2\infty} }$$ = $${ \lq\lq \ (\infty)^2 \ '' \over e^{\infty} }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule. Differentiate the top and the bottom using the chain rule.) truein

= $${ \lim_{x \to \infty} \ { 2(\ln x) \cdot (1/x) \over 2e^{2x} } }$$

truein

= $${ \lim_{x \to \infty} \ { \ln x \over xe^{2x} } }$$

truein

= $${ \lq\lq \ \ln(\infty) \ '' \over (\infty)e^{2\infty} }$$ = $${ \lq\lq \ \infty \ '' \over (\infty)e^{\infty} }$$ = $${ \lq\lq \ \infty \ '' \over (\infty)(\infty) }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule again.) truein

= $${ \lim_{x \to \infty} \ { 1/x \over x \cdot 2e^{2x}+(1)e^{2x} } }$$

truein

= $${ \lim_{x \to \infty} \ { {1 \over x} \cdot { 1 \over 2xe^{2x}+ e^{2x} } } }$$

truein

= $${ \lim_{x \to \infty} \ { {1 \over 2x^2e^{2x}+xe^{2x} } } }$$

truein

= $${ 1 \over 2(\infty)^2e^{2\infty}+(\infty)e^{2\infty} }$$

truein

= $${ 1 \over \infty + \infty }$$

truein

= $${ 1 \over \infty }$$ truein

= $${ 0}$$ .

truein truein SOLUTION 15: $${ \lim_{x \to \infty} \ { 3x+2^x \over 2x+3^x } }$$ = $${ \lq\lq \ 3(\infty)+2^{\infty} \ '' \over 2(\infty)+3^{\infty} }$$ = $${ \lq\lq \ \infty + \infty \ '' \over \infty + \infty }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule. Differentiate the top and the bottom.) truein

= $${ \lim_{x \to \infty} \ { 3+2^x \ln 2 \over 2+3^{x} \ln 3 } }$$

truein

= $${ \lq\lq \ 3+2^{\infty} \ln 2 \ '' \over 2+3^{\infty} \ln 3 }$$ = $${ \lq\lq \ 3 + \infty \ '' \over 2 + \infty }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule again.)

truein

= $${ \lim_{x \to \infty} \ {0+2^{x} \ln 2 \cdot \ln 2 \over 0+3^{x} \ln 3 \cdot \ln 3 } }$$

truein

= $${ \lim_{x \to \infty} \ {2^{x} (\ln 2)^2 \over 3^{x} (\ln 3)^2 } }$$

truein

= $${ {2^{\infty} (\ln 2)^2 \over 3^{\infty} (\ln 3 )^2 } }$$ = $${ {(\infty) (\ln 2)^2 \over (\infty) (\ln 3 )^2 } }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Applying Theorem 2 for l'Hopital's Rule again will lead to $${ \lim_{x \to \infty} \ {2^{x} (\ln 2)^3 \over 3^{x} (\ln 3)^3 } }$$. Stop and think. This is going nowhere. There is an easy way out.)

truein

= $${ \lim_{x \to \infty} \ {2^{x} (\ln 2)^2 \over 3^{x} (\ln 3)^2 } }$$

truein

= $${ \lim_{x \to \infty} \ {2^{x} \over 3^{x} } \cdot { (\ln 2)^2 \over (\ln 3 )^2 } }$$

truein

= $${ \lim_{x \to \infty} \ \Big( {2 \over 3} \Big)^x \cdot { (\ln 2)^2 \over (\ln 3 )^2 } }$$

truein

= $${ \Big( {2 \over 3} \Big)^{\infty} \cdot { (\ln 2)^2 \over (\ln 3 )^2 } }$$

truein

= $${ (0) { (\ln 2)^2 \over (\ln 3 )^2 } }$$

truein

= $${ 0}$$ .

truein truein SOLUTION 16: $${ \lim_{x \to \infty} \ { e^x + {2/x} \over e^x + {5/x } } }$$ = $${ \lq\lq \ e^{\infty} + 2/ \infty \ '' \over e^{\infty} + 5/ \infty }$$ = $${ \lq\lq \ \infty + 0 \ '' \over \infty + 0 }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Apply Theorem 2. Differentiate the top and the bottom.) truein

= $${ \lim_{x \to \infty} \ { e^x - 2/x^2 \over e^x - 5/x^2 } }$$ = $${ \lq\lq \ \infty - 0 \ '' \over \infty - 0 }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Applying Theorem 2 for l'Hopital's Rule again will get us nowhere. Stop and think. Go back to the original problem and rewrite it.) truein

= $${ \lim_{x \to \infty} \ { { e^x + {2/x} \over e^x + {5/x } } \cdot { x \over x } } }$$

truein

= $${ \lim_{x \to \infty} \ { xe^x + 2 \over xe^x + 5 } }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule. Differentiate the top and bottom using the product rule.)

truein

= $${ \lim_{x \to \infty} \ { xe^x + (1)e^x \over xe^x + (1)e^x } }$$

truein

= $${ \lim_{x \to \infty} \ { 1 } }$$

truein

= $${ { 1 } }$$ .

truein truein SOLUTION 17: $${ \lim_{x \to \infty} \ ( \sqrt{ x^2+1 } - \sqrt{ x+1 } ) }$$ = $\lq\lq \ \infty - \infty \ ''$

truein (Rewrite the expression by using a conjugate to circumvent this indeterminate form.) truein

= $${ \lim_{x \to \infty} \ ( \sqrt{ x^2+1 } - \sqrt{ x+1 } ) \cdot {\sqrt{ x^2+1 } + \sqrt{ x+1 } \over \sqrt{ x^2+1 } + \sqrt{ x+1 } } }$$

truein (Recall that $(A-B)(A+B)=A^2-B^2$.) truein

= $${ \lim_{x \to \infty} \ { (x^2+1) - (x+1) \over { \sqrt{ x^2+1 } + \sqrt{ x+1 } } } }$$

truein

= $${ \lim_{x \to \infty} \ { x^2 - x \over { \sqrt{ x^2+1 } + \sqrt{ x+1 } } } }$$

truein

= $${ \lim_{x \to \infty} \ { x(x-1) \over \sqrt{ x^2+1 } + \sqrt{ x+1 } } }$$ = $${ \lq\lq \ (\infty)(\infty) \ '' \over \infty + \infty }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule. Use the chain rule in the bottom.) truein

= $${ \lim_{x \to \infty} \ { 2x-1 \over (1/2)(x^2+1)^{-1/2} (2x) + (1/2)(x+1)^{-1/2} } }$$

truein

= $${ \lim_{x \to \infty} \ { 2x-1 \over { x \over \sqrt{ x^2+1 } } + { 1 \over 2\sqrt{ x+1 } } } }$$

truein

= $${ \lim_{x \to \infty} \ { 2x-1 \over { \sqrt{x^2} \over \sqrt{ x^2+1 } } + { 1 \over 2\sqrt{ x+1 } } } }$$

truein

= $${ \lim_{x \to \infty} \ { 2x-1 \over { \sqrt{x^2 \over x^2+1 } } + { 1 \over 2\sqrt{ x+1 } } } }$$

truein

= $${ \lim_{x \to \infty} \ { 2x-1 \over { \sqrt{ {x^2 \over x^2+1} \cdot { 1/x^2 \over 1/x^2} } } + { 1 \over 2\sqrt{ x+1 } } } }$$

truein

= $${ \lim_{x \to \infty} \ { 2x-1 \over { \sqrt{ {1 \over 1+1/x^2} } } + { 1 \over 2\sqrt{ x+1 } } } }$$

truein

= $${ { 2 ( \infty)-1 \over { \sqrt{ {1 \over 1+1/{\infty}^2} } } + { 1 \over 2\sqrt{ \infty } } } }$$

truein

= $${ \infty \over \sqrt{1 \over 1+0} + {1 \over \infty} }$$

truein

= $${ \infty \over 1 + {0} }$$

truein

= $${ \infty }$$ (The limit does not exist.) .

truein truein SOLUTION 18: $${ \lim_{x \to 0^+} \ { x \cdot \ln x} }$$ = $\lq\lq \ 0 \cdot \ln 0 \ ''$ = $\lq\lq \ 0 \cdot (-\infty) \ ''$

truein (This is an indeterminate form. ``Flip" $x$ so that l'Hopital's Rule can be applied.) truein

= $${ \lim_{x \to 0^+} \ { \ln x \over 1/x } }$$ = $${ { \lq\lq \ \ln 0 \ '' \over 1/0^+ } }$$ = $${ \lq\lq \ - \infty \ '' \over \infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule. Differentiate the top and the bottom separately.) truein

= $${ \lim_{x \to 0^+} \ { 1/x \over -1/x^2} }$$

truein

= $${ \lim_{x \to 0^+} \ { 1 \over x } \cdot { x^2 \over -1 } }$$

truein

= $${ \lim_{x \to 0^+ } \ (-x) }$$

truein

= $0$ .

truein truein SOLUTION 19: $${ \lim_{x \to 0^+} \ { x \cdot ( \ln x )^2 } }$$ = $\lq\lq \ 0 \cdot ( \ln 0 )^2 \ ''$ = $\lq\lq \ 0 \cdot (-\infty)^2 \ ''$ = $\lq\lq \ 0 \cdot \infty \ ''$

truein (This is an indeterminate form. ``Flip" $x$ so that l'Hopital's Rule can be applied.) truein

= $${ \lim_{x \to 0^+} \ { (\ln x)^2 \over 1/x } }$$ = $${ { \lq\lq \ (\ln 0)^2 \ '' \over 1/0^2 } }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule.) truein

= $${ \lim_{x \to 0^+} \ {2 \ln x \cdot 1/x \over -1/x^2} }$$

truein

= $${ \lim_{x \to 0^+} \ {2 \ln x \over -1/x} }$$ = $${ \lq\lq \ -\infty \ '' \over -\infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule again.) truein

= $${ \lim_{x \to 0^+} \ { { 2/x } \over { 1/x^2 } } }$$

truein

= $${ \lim_{x \to 0^+ } \ 2x }$$

truein

= $0$ .

truein truein SOLUTION 20: $${ \lim_{x \to 0^+} \ { \ln x \cdot \tan x } }$$ = $${ \lq\lq \ \ln 0 \cdot \tan 0 \ '' }$$ = $${ \lq\lq \ (-\infty) \cdot 0 \ '' }$$

truein (Rewrite the expression to circumvent this indeterminate form. Recall that $\tan x = 1/ \cot x$.) truein

= $${ \lim_{x \to 0^+} \ { \ln x \over \cot x } }$$ = $${ \lq\lq \ \ln 0 \ '' \over \cot 0 }$$ = $${ \lq\lq \ - \infty \ '' \over \infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule.) truein

= $${ \lim_{x \to 0^+} \ { 1/x \over -\csc^2 x } }$$

truein (Recall that $\csc x = 1/ \sin x$.) truein

= $${ \lim_{x \to 0^+} \ { 1/x \over -1/ \sin^2 x } }$$

truein

= $${ \lim_{x \to 0^+} \ { 1 \over x} \cdot { \sin^2 x \over -1 } }$$

truein

= $${ \lim_{x \to 0^+} \ { - \sin^2 x \over x } }$$ = $${{ \lq\lq \ - \sin^2 0 \ '' \over 0 } }$$ = $${{\lq\lq \ (0)^2 \ '' \over 0 } }$$ = $${{\lq\lq \ 0 \ '' \over 0 } }$$

truein (Apply Theorem 1 for l'Hopital's Rule.)

truein

= $${ \lim_{x \to 0^+} \ { - 2 \sin x \cos x \over 1 } }$$

truein

= $${ - 2 \sin 0 \cdot \cos 0 }$$

truein

= $${ - 2 (0) (1) }$$

truein

= $${ 0}$$ .

truein truein SOLUTION 21: $${ \lim_{x \to 0^+} \ x^{ \ \sin x } }$$ = $${ \lq\lq \ 0^{ \ \sin 0 } \ '' }$$ = $${ \lq\lq \ 0^{ \ 0 } \ '' }$$

truein (Rewrite the problem to circumvent this indeterminate form. Recall that $${ e^{\ln z} = z }.$$) truein

= $${ \lim_{x \to 0^+ } \ e^{ \ \ln x^{ \ \sin x } } }$$

truein

= $${ \lim_{x \to 0^+ } \ e^{ \ { \sin x \cdot \ln x } } }$$

truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ { \sin x \ln x }}}}$$ = $${ \ e^ { \ \displaystyle{ \lq\lq \ \sin 0 \cdot \ln 0 \ '' } } }$$ = $${ \ e^ { \ \displaystyle{ \lq\lq \ 0 \cdot (-\infty) \ '' } } }$$

truein (``Flip" $\sin x$ to circumvent this indeterminate form.) truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ { \ln x \over {1/ \sin x} } } } }$$

truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ { \ln x \over \csc x } } } }$$ = $${ \ e^ { \ \displaystyle{ \lq\lq \ \ln 0 / \csc 0 \ '' } } }$$ = $${ \ e^ { \ \displaystyle{ { \lq\lq -\infty / \infty \ '' } } } }$$

truein (Apply Theorem 2 for l'Hopital's Rule.) truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ { {1/x} \over {-\csc x \cot x} } } } }$$

truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ { {1/x} \over {-1 \over \sin x} \cdot { \cos x \over \sin x } } } } }$$

truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ { -\sin^2 x \over x \cos x } } } }$$ = $${ \ e^ { \ \displaystyle{ \lq\lq \ -\sin^2 0 \ '' \over 0 \cdot \cos 0 } } }$$ = $${ \ e^ { \ \displaystyle{ \lq\lq \ -(0)^2 \ '' \over (0)(1) } } }$$ = $e^{\displaystyle{ \lq\lq \ 0/0 \ '' }}$

truein (Apply Theorem 1 for l'Hopital's Rule.) truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to 0^+ } \ { -2\sin x \cdot \cos x \over \cos x - x \sin x } } } }$$

truein

= $${ \ e^ { \ \displaystyle{ { -2 \sin 0 \cdot \cos 0 \over \cos 0 - 0 \cdot \sin 0 } } } }$$

truein

= $${ \ e^ { \ \displaystyle{ { -2(0)(1) \over (1) - (0)(0) } } } }$$

truein

= $${ \ e^ { \ \displaystyle{ { 0} } } }$$

truein

= $1$ .

truein truein SOLUTION 22: $${ \lim_{x \to \infty} \ (1 + 3/x)^x }$$ = $${ \lq\lq \ (1+0)^{ \ \infty } \ '' }$$ = $${ \lq\lq \ 1^{ \ \infty } \ '' }$$

truein (Rewrite the problem to circumvent this indeterminate form. Recall that $${ e^{\ln z} = z }.$$) truein

= $${ \lim_{x \to \infty } \ e^{ \ \ln (1 + 3/x)^x } }$$

truein

= $${ \lim_{x \to \infty } \ e^{ \ x \cdot \ln (1 + 3/x) } }$$

truein

= $${ e^{ \ \displaystyle{ \lim_{x \to \infty } \ x \cdot \ln (1 + 3/x) } } }$$ = $${ \ e^ { \ \displaystyle{ \lq\lq \ \infty \cdot \ln 1 \ '' } } }$$ = $${ \ e^ { \ \displaystyle{ \lq\lq \ \infty \cdot 0 \ '' } } }$$

truein (``Flip" $x$ to circumvent this indeterminate form.) truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to \infty } \ { \ln (1 + 3/x) \over {1/x} } } } }$$ = $${ \ e^ { \ \displaystyle{ { \lq\lq \ \ln 1 / (1/ \infty) \ '' } } } }$$ = $${ \ e^ { \ \displaystyle{ { \lq\lq \ 0 / 0 \ '' } } } }$$

truein (Apply Theorem 1 for l'Hopital's Rule. Recall that $${ D \{ \ln f(x) \} = \frac{1}{f(x)} \cdot f'(x) }.$$) truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to \infty } \ { { 1 \over 1 + 3/x } \cdot {-3/x^2} \over { -1/x^2 } } } } }$$

truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to \infty } \ { { 3 \over 1 + 3/x } } } } }$$

truein

= $${ \ e^ { \ \displaystyle{{3}/(1+0) } } }$$

truein

= $${ \ e^ { \ \displaystyle{ 3 } } }$$ .

truein truein SOLUTION 23: $${ \lim_{x \to 0} \ (1 - x)^{1/x} }$$ = $${ \lq\lq \ (1-0)^{ \ \pm \infty } \ '' }$$ = $${ \lq\lq \ 1^{ \ \pm \infty } \ '' }$$

truein (Rewrite the problem to circumvent this indeterminate form. Recall that $${ e^{\ln z} = z }.$$) truein

= $${ \lim_{x \to 0 } \ e^{ \ \displaystyle \ln (1 - x)^{1/x} } }$$

truein

= $${ \lim_{x \to 0 } \ e^{ \ \displaystyle {1/x} \cdot \ln (1-x) } }$$

truein

= $${ e^{ \ \displaystyle{ \lim_{x \to 0 } \ {\ln (1-x) \over x }}}}$$ = $${ \ e^ { \ \displaystyle{ \lq\lq \ { \ln 1/0} \ '' } } }$$ = $${ \ e^ { \ \displaystyle{ \lq\lq \ { 0/0}\ '' } } }$$

truein (Apply Theorem 1 for l'Hopital's Rule.) truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to 0} \ { { 1 \over 1-x } \cdot (-1) \over {1} } } } }$$

truein

= $${ \ e^ { \ \displaystyle{ { { -1/(1-0) } } } } }$$

truein

= $${ \ e^ { \ \displaystyle{-1} } }$$

truein

= $${ 1 \over e }$$ .

truein truein SOLUTION 24: $${ \lim_{x \to 0^+} \ (\tan x)^{x^2} }$$ = $${ \lq\lq \ (\tan 0)^{ \ 0 } \ '' }$$ = $${ \lq\lq \ 0^{ \ 0 } \ '' }$$

truein (Rewrite the problem to circumvent this indeterminate form. Recall that $${ e^{\ln z} = z }.$$) truein

= $${ \lim_{x \to 0^+ } \ e^{ \ \displaystyle \ln (\tan x)^{x^2} } }$$

truein

= $${ \lim_{x \to 0^+ } \ e^{ \ \displaystyle x^2 \cdot \ln (\tan x) } }$$

truein

= $${ e^{ \ \displaystyle{ \lim_{x \to 0^+ } \ x^2 \cdot \ln (\tan x) } } }$$

truein

= $${ \ e^ { \ \displaystyle{ \lq\lq \ 0 \cdot \ln (\tan 0) \ '' } } }$$ = $${ \ e^ { \ \displaystyle{ \lq\lq \ 0 \cdot \ln 0 \ '' } } }$$ = $${ \ e^ { \ \displaystyle{ \lq\lq \ 0 \cdot (-\infty) \ '' } } }$$

truein (``Flip" $x^2$ to circumvent this indeterminate form.) truein

= $${ e^{ \ \displaystyle{ \lim_{x \to 0^+ } \ { \ln (\tan x) \over 1/x^2 } } } }$$

truein

= $${ \ e^ { \ \displaystyle{ \lq\lq \ \ln (\tan 0)/(1/0^2) \ '' } } }$$ = $${ \ e^ { \ \displaystyle{ \lq\lq \ \ln 0 / \infty \ '' } } }$$ = $${ \ e^ { \ \displaystyle{ \lq\lq \ -\infty / \infty \ '' } } }$$

truein (Apply Theorem 2 for l'Hopital's Rule.) truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to 0^+} \ { { (1/\tan x) } \cdot \sec^2 x \over -2/x^3 } } } }$$

truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to 0^+} \ { { \sec^2 x \over \tan x } \cdot {x^3 \over -2} } } } }$$

truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to 0^+} \ { { -x^3 \sec^2 x \over 2 \tan x } } } } }$$

truein

= $${ \ e^ { \ \displaystyle{ { { \lq\lq \ 0 \cdot \sec^2 0 \ '' \over 2 \tan 0 } } } } }$$ = $${ \ e^ { \ \displaystyle{ { { \lq\lq \ (0)(1)^2 \ '' \over 2 (0) } } } } }$$ = $${ \ e^ { \ \displaystyle{ \lq\lq \ 0/0 \ '' } } }$$

truein (Apply Theorem 1 for l'Hopital's Rule.) truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to 0^+} \ { { -x^3 \cdot 2 \sec x \cdot \sec x \tan x -3x^2 \sec^2 x \over 2 \sec^2 x }}}}}$$

truein

= $${ \ e^ { \ \displaystyle{ \lim_{x \to 0^+} \ { { -2x^3 \sec^2 x \tan x -3x^2 \sec^2 x \over 2 \sec^2 x }}}}}$$

truein

= $${ \ e^ { \ \displaystyle{ { { 0 \cdot \sec^2 0 \cdot \tan 0 - 0 \cdot \sec^2 0 \over 2 \sec^2 0 }}}}}$$

truein

= $${ \ e^ { \ \displaystyle{ { { (0)(1)^2 (0) - (0)(1)^2 \over 2 (1)^2 }}}}}$$

truein

= $${ \ e^ { \ \displaystyle{0/2} }}$$

truein

= $e^{ \ \displaystyle{0}}$

truein

= $1$ .

truein truein SOLUTION 25: $${ \lim_{x \to \infty} \ x^{ \ 1/ \sqrt{x}} }$$ = $${ \lq\lq \ \infty^{ \ 1/ \infty } \ '' }$$ = $${ \lq\lq \ \infty^{ \ 0} \ '' }$$

truein (Rewrite the problem to circumvent this indeterminate form. Recall that $${ e^{\ln z} = z }.$$) truein

= $${ \lim_{x \to \infty } \ e^{ \ \displaystyle \ln x^{1/ \sqrt{x}} } }$$

truein

= $${ \lim_{x \to \infty } \ e^{ \ \displaystyle (1/ \sqrt{x}) \ln x } }$$

truein

= $${ \lim_{x \to \infty } \ e^{ \ \displaystyle \ln x / \sqrt{x} } }$$

truein

= $${ e^{ \ \displaystyle{ \lim_{x \to \infty } \ { \ln x / \sqrt{x} } } } }$$ = $${ \ e^{ \ \displaystyle{ \lq\lq \ \ln (\infty) / \sqrt{\infty} \ '' } } }$$ = $${ \ e^{ \ \displaystyle{ \lq\lq \ \infty / \infty \ '' } } }$$

truein (Apply Theorem 2 for l'Hopital's Rule.) truein

= $${ e^{ \ \displaystyle{ \lim_{x \to \infty } \ { (1/x) / (1/2\sqrt{x}) } } } }$$

truein

= $${ e^{ \ \displaystyle{ \lim_{x \to \infty } \ { 2\sqrt{x} / x } } } }$$

truein

= $${ e^{ \ \displaystyle{ \lim_{x \to \infty } \ { 2 / \sqrt{x} } } } }$$

truein

= $${ e^{ \ \displaystyle{ \lq\lq \ { 2 / \sqrt{\infty} \ '' } } } }$$

truein

= $${ e^{ \ \displaystyle{ \lq\lq \ { 2 / \infty \ '' } } } }$$

truein

= $${ e^{ \ \displaystyle{ 0 } } }$$

truein

= $${1 }$$ .

truein truein SOLUTION 26: $${ \lim_{x \to \infty} \ ( \ln x)^{1/x} }$$ = $${ \lq\lq \ \ln(\infty)^{ \ 1/ \infty } \ '' }$$ = $${ \lq\lq \ \infty^{ \ 0} \ '' }$$

truein (Rewrite the problem to circumvent this indeterminate form. Recall that $${ e^{\ln z} = z }.$$) truein

= $${ \lim_{x \to \infty } \ e^{ \ \displaystyle \ln x^{1/x} } }$$

truein

= $${ \lim_{x \to \infty } \ e^{ \ \displaystyle (1/x) \ln x } }$$

truein

= $${ \lim_{x \to \infty } \ e^{ \ \displaystyle \ln x / x } }$$

truein

= $${ e^{ \ \displaystyle{ \lim_{x \to \infty } \ { \ln x / x } } } }$$ = $${ \ e^{ \ \displaystyle{ \lq\lq \ \ln (\infty) / \sqrt{\infty} \ '' } } }$$ = $${ \ e^{ \ \displaystyle{ \lq\lq \ \infty / \infty \ '' } } }$$

truein (Apply Theorem 2 for l'Hopital's Rule.) truein

= $${ e^{ \ \displaystyle{ \lim_{x \to \infty } \ { (1/x) / 1 } } } }$$

truein

= $${ e^{ \ \displaystyle{ \lq\lq \ 1/\infty \ '' } } }$$

truein

= $${ e^{ \ \displaystyle{ 0 } } }$$

truein

= $${1 }$$ .

truein truein SOLUTION 27: $${ \lim_{x \to 0^+} \ x^{x^x} }$$ = $${ \lq\lq \ 0^{ \ 0^{ \ 0}} \ '' }$$

truein (Rewrite the problem to circumvent this indeterminate form. Recall that $${ e^{\ln z} = z }.$$) truein

= $${ \lim_{x \to 0^+ } \ e^{ \ \displaystyle \ln x^{x^x} } }$$

truein

= $${ \lim_{x \to 0^+ } \ e^{ \ \displaystyle x^x \ln x } }$$

truein

= $${ e^{ \ \displaystyle{ \lim_{x \to 0^+ } \ x^x \ln x }}}$$

truein

= $${ e^{ \ \displaystyle{ (\lim_{x \to 0^+ } \ x^x)( \lim_{x \to 0^+ } \ \ln x) }}}$$

truein (See Example 4 in the material preceeding this problem set.) truein

= $${ e^{ \ \displaystyle{ \lq\lq \ (1)(\ln 0) \ '' }}}$$ truein

= $${ e^{ \ \displaystyle{ \lq\lq \ (1)(-\infty) \ '' }}}$$

truein

= $${ \lq\lq \ 1/ e^{\infty} \ '' }$$

truein

= $${ \lq\lq \ 1/ \infty \ '' }$$

truein

= $${ 0}$$ .

truein truein SOLUTION 28: a.) Start with $${ -1 \le \sin x \le +1 }$$ , truein so that

truein $${ 3x-1 \le 3x+\sin x \le 3x+1 }$$

truein and (Assume that $x>0$.)

truein $${ { 3x-1 \over 2x } \le { 3x+\sin x \over 2x } \le { 3x+1 \over 2x } }$$ .

truein Then $${ \lim_{x \to \infty} \ { 3x-1 \over 2x } }$$ = $${ \lim_{x \to \infty} \ { 3x-1 \over 2x } \cdot { 1/x \over 1/x } }$$

truein

= $${ \lim_{x \to \infty} \ { 3x/x -1/x \over 2x/x } }$$

truein

= $${ \lim_{x \to \infty} \ { 3 -1/x \over 2 } }$$

truein

= $${ 3 - 0 \over 2 }$$

truein

= $${ 3 \over 2 }$$ .

truein Similarly, $${ \lim_{x \to \infty} \ { 3x+1 \over 2x } }$$ = $${ 3 \over 2 }$$ .

truein Since

truein $${ \lim_{x \to \infty} \ { 3x-1 \over 2x } }$$ = $${ 3 \over 2 }$$ = $${ \lim_{x \to \infty} \ { 3x+1 \over 2x } }$$ ,

truein it follows from the Squeeze Principle that

truein $${ \lim_{x \to \infty} \ { 3x+ \sin x \over 2x } }$$ = $${ 3 \over 2 }$$ .

truein truein SOLUTION 28: b.) $${ \lim_{x \to \infty} \ { 3x+ \sin x \over 2x } }$$ = $${ \lq\lq \ \infty \ '' \over \infty }$$

truein (Apply Theorem 2 for l'Hopital's Rule.) truein

= $${ \lim_{x \to \infty} \ { 3+ \cos x \over 2 } }$$

truein which does not exist since truein

$${ -1 \le \cos x \le +1 }$$ .

truein truein SOLUTION 28: c.) The answers to parts a.) and b.) tell us that l'Hopital's Rule may give us a wrong answer if the answer is `` does not exist." We can only be sure that l'Hopital's Rule gives us the correct answer if the answer is finite, $+ \infty$, or $- \infty$ .