The following problems involve the use of l'Hopital's Rule. It is used to circumvent the common indeterminate forms $ \frac{ "0" }{ 0 } $ and $ \frac{"\infty" }{ \infty } $ when computing limits. There are numerous forms of l"Hopital's Rule, whose verifications require advanced techniques in calculus, but which can be found in many calculus books. This link will show you the plausibility of l'Hopital's Rule. Following are two of the forms of l'Hopital's Rule.
THEOREM 1 (l'Hopital's Rule for zero over zero): Suppose that $ \displaystyle{ \lim_{x \rightarrow a} f(x) =0 } $, $ \displaystyle{ \lim_{x \rightarrow a} g(x) =0 } $, and that functions $f$ and $g$ are differentiable on an open interval $I$ containing $a$. Assume also that $ g'(x) \ne 0$ in $I$ if $x \ne a$. Then $$ \displaystyle{ \lim_{x \rightarrow a} \frac{f(x)}{g(x)} } = \displaystyle{ \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} } $$ so long as the limit is finite, $+\infty$, or $-\infty$. Similar results hold for $ x \rightarrow \infty $ and $ x \rightarrow - \infty $.
THEOREM 2 (l'Hopital's Rule for infinity over infinity): Assume that functions $f$ and $g$ are differentiable for all $x$ larger than some fixed number. If $ \displaystyle{ \lim_{x \rightarrow a} f(x) = \infty } $ and $ \displaystyle{ \lim_{x \rightarrow a} g(x) = \infty } $, then $$ \displaystyle{ \lim_{x \rightarrow a} \frac{f(x)}{g(x)} } = \displaystyle{ \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} } $$ so long as the limit is finite, $+\infty$, or $-\infty$. Similar results hold for $ x \rightarrow \infty $ and $ x \rightarrow - \infty $.
In both forms of l'Hopital's Rule it should be noted that you are required to differentiate (separately) the numerator and denominator of the ratio if either of the indeterminate forms $ \frac{ "0" }{ 0 } $ or $ \frac{"\infty" }{ \infty } $ arises in the computation of a limit. Do not confuse l'Hopital's Rule with the Quotient Rule for derivatives. Here is a simple illustration of Theorem 1.
EXAMPLE 1: $$ \displaystyle{ \lim_{x \rightarrow 2} \frac{x-2}{x^2-4} = \frac{" 2-2" }{ (2)^2-4 } = \frac{ "0" }{ 0 } } $$ (Now apply Theorem 1. Differentiate top and bottom separately.) $$ = \displaystyle{ \lim_{x \rightarrow 2} \frac{1-0}{2x-0} } $$ $$ = \displaystyle{ \lim_{x \rightarrow 2} \frac{1}{2x} } $$ $$ = \frac{1}{2(2)} $$ $$ = \frac{1}{4} $$ Here is a simple illustration of Theorem 1.
EXAMPLE 2: $$ \displaystyle{ \lim_{x \rightarrow \infty} \frac{2x+7}{3x^2-5} = \frac{"\infty" }{ \infty } = \frac{ "0" }{ 0 } } $$ (Now apply Theorem 2. Differentiate top and bottom separately.) $$ = \displaystyle{ \lim_{x \rightarrow \infty} \frac{2+0}{6x-0} } $$ $$ = \displaystyle{ \lim_{x \rightarrow \infty} \frac{1}{3x} } $$ $$ = \frac{"1"}{\infty} $$ $$ = 0 $$
Indeterminate forms besides $ \frac{ "0" }{ 0 } $ and $ \frac{"\infty" }{ \infty } $ include $ "0 \cdot \infty $, $ "\infty - \infty" $, $ "1^{\infty}" $, $ "0^{0}" $, and $ "\infty^{0}" $. These forms also arise in the computation of limits and can often be algebraically transformed into the form $ \frac{ "0" }{ 0 } $ or $ \frac{"\infty" }{ \infty } $, so that l'Hopital's Rule can be applied. Following are two examples of such transformations. The second example uses the fact that $ y=e^x $ and $ y=\ln x $ are inverse functions, so that $ z= e^{\ln z} $ for all $ z>0 $ and $ \ln z^m=m \ln z$ for all $ z>0 $ and any $ m$.
EXAMPLE 3: $$ \displaystyle{ \lim_{x \rightarrow 0^{+}} \sqrt{x} \cdot \ln x } = " 0 \cdot -\infty"$$ (Circumvent this indeterminate form by "flipping" $ \sqrt{x} $.) $$ = \displaystyle{ \lim_{x \rightarrow 0^{+}} \frac{\ln x}{1/\sqrt{x}} } = \frac{ "- \infty" }{ \infty } $$ (Now use Theorem 2 for l'Hopital's Rule.) $$ = \displaystyle{ \lim_{x \rightarrow 0^{+}} \frac{1/x}{-1/2x^{3/2}} } $$ $$ = \displaystyle{ \lim_{x \rightarrow 0^{+}} -2 \sqrt{x} } $$ $$ = -2 \sqrt{0} $$ $$ = -2 (0) $$ $$ = 0 $$
EXAMPLE 4: $$ \displaystyle{ \lim_{x \rightarrow 0^{+}} x^x } = " 0^0" $$ (Use the fact that $ z=e^{\ln z} $.) $$ = \displaystyle{ \lim_{x \rightarrow 0^{+}} e^{ \ln {x^{x}} } } $$ (Use the fact that $ \ln z^m=m \ln z$ .) $$ = \displaystyle{ \lim_{x \rightarrow 0^{+}} e^{ x \ln x } } $$ (This next step uses the fact that $y=e^{x \ln x} $ is a continuous function.) $$ = \displaystyle{ e^{\lim_{x \rightarrow 0^{+}} { x \ln x }} } $$ $$ = \displaystyle{ e^{ "0 \cdot (-\infty)"} } $$ (Circumvent this indeterminate form by "flipping" $ x $.) $$ = \displaystyle{ e^{\lim_{x \rightarrow 0^{+}} { \frac{ \ln x }{1/x} } } } $$ (Now apply Theorem 2 for l'Hopita's Rule.) $$ = \displaystyle{ e^{\lim_{x \rightarrow 0^{+}} { \frac{ 1/x }{-1/x^2} } } } $$ $$ = \displaystyle{ e^{\lim_{x \rightarrow 0^{+}} { (-x) } } } $$ $$ = e^0 $$ $$ = 1 $$
In the list of limit problems which follows, most problems are average and a few are somewhat challenging. In some cases there may be methods other than l'Hopital's Rule that could be used to compute the given limit. Nonetheless, l'Hopital's Rule will be used whenever applicable in this problem set.
Click HERE to see a detailed solution to problem 1.
PROBLEM 2: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow 4} \frac{ x-4 }{ \sqrt{x} - 2 } } $ .Click HERE to see a detailed solution to problem 2.
PROBLEM 3: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow 0} \frac{ \sin x }{ x } } $ .Click HERE to see a detailed solution to problem 3.
PROBLEM 4: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow 0} \frac{ 3^x - 2^x }{ x^2-x } } $ .Click HERE to see a detailed solution to problem 4.
PROBLEM 5: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow 3} \frac{ 1/x - 1/3 }{ x^2-9 } } $ .Click HERE to see a detailed solution to problem 5.
PROBLEM 6: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow 0} \frac{ x \tan x }{ \sin 3x } } $ .Click HERE to see a detailed solution to problem 6.
PROBLEM 7: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow 0} \frac{ \arcsin 4x }{ \arctan 5x } } $ .Click HERE to see a detailed solution to problem 7.
PROBLEM 8: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow 0} \frac{ \sin x^2 }{ x \tan x } } $ .Click HERE to see a detailed solution to problem 8.
PROBLEM 9: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow 0} \frac{ x^2 e^x}{ \tan^2 x } } $ .Click HERE to see a detailed solution to problem 9.
PROBLEM 10: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow 0} \frac{ e^{-1/x^2}}{ x^2 } } $ .Click HERE to see a detailed solution to problem 10.
PROBLEM 11: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow \infty} \frac{ e^x }{ 5x+200 } } $ .Click HERE to see a detailed solution to problem 11.
PROBLEM 12: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow \infty} \frac{ 3+\ln x }{ x^2+7 } } $ .Click HERE to see a detailed solution to problem 12.
PROBLEM 13: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow \infty} \frac{ x^2+3x-10 }{ 7x^2-5x+4 } } $ .Click HERE to see a detailed solution to problem 13.
PROBLEM 14: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow \infty} \frac{ ( \ln x)^2 }{ e^{2x} } } $ .Click HERE to see a detailed solution to problem 14.
PROBLEM 15: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow \infty} \frac{ 3x+2^x }{ 2x+3^x } } $ .Click HERE to see a detailed solution to problem 15.
PROBLEM 16: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow \infty} \frac{ e^x + 2/x }{ e^x + 5/x } } $ .Click HERE to see a detailed solution to problem 16.
PROBLEM 17: Compute $ \ \ \displaystyle{ \lim_{x \rightarrow \infty} ( \sqrt{x^2+1} - \sqrt{x+1} ) } $ . $ \ \ $ (Recall that $ "\infty - \infty" $ is an indeterminate form. HINT: First use a conjugate.)Click HERE to see a detailed solution to problem 17.