SOLUTION 17: $$ \displaystyle{ \lim_{x \to \infty} \
( \sqrt{ x^2+1 } - \sqrt{ x+1 } ) } = `` \ \infty - \infty \ " $$
(Rewrite the expression by using a conjugate to circumvent this indeterminate form.)
$$ \displaystyle{ \lim_{x \to \infty} \
( \sqrt{ x^2+1 } - \sqrt{ x+1 } ) \cdot {\sqrt{ x^2+1 } + \sqrt{ x+1 } \over
\sqrt{ x^2+1 } + \sqrt{ x+1 } } } $$
(Recall that $ (A-B)(A+B)=A^2-B^2 $.)
$$ = \displaystyle{ \lim_{x \to \infty} \ { (x^2+1) - (x+1) \over
{ \sqrt{ x^2+1 } + \sqrt{ x+1 } } } } $$
$$ = \displaystyle{ \lim_{x \to \infty} \ { x^2 - x \over { \sqrt{ x^2+1 } + \sqrt{ x+1 } } } } = \displaystyle{ \lim_{x \to \infty} \ { x(x-1) \over \sqrt{ x^2+1 } + \sqrt{ x+1 } } } = \displaystyle{ `` \ (\infty)(\infty) \ " \over \infty + \infty }
= \displaystyle{ `` \ \infty \ " \over \infty } $$
(Apply Theorem 2 for l'Hopital's Rule. Use the chain rule in the bottom.)
$$ = \displaystyle{ \lim_{x \to \infty} \
{ 2x-1 \over (1/2)(x^2+1)^{-1/2} (2x) + (1/2)(x+1)^{-1/2} } } $$
$$ = \displaystyle{ \lim_{x \to \infty} \
{ 2x-1 \over { x \over \sqrt{ x^2+1 } } + { 1 \over 2\sqrt{ x+1 } } } } $$
$$ = \displaystyle{ \lim_{x \to \infty} \
{ 2x-1 \over { \sqrt{x^2} \over \sqrt{ x^2+1 } } + { 1 \over 2\sqrt{ x+1 } } } } $$
$$ = \displaystyle{ \lim_{x \to \infty} \
{ 2x-1 \over { \sqrt{x^2 \over x^2+1 } } + { 1 \over 2\sqrt{ x+1 } } } } $$
$$ = \displaystyle{ \lim_{x \to \infty} \
{ 2x-1 \over { \sqrt{ {x^2 \over x^2+1} \cdot { 1/x^2 \over 1/x^2} } } + { 1 \over 2\sqrt{ x+1 } } } } $$
$$ = \displaystyle{ \lim_{x \to \infty} \
{ 2x-1 \over { \sqrt{ {1 \over 1+1/x^2} } } + { 1 \over 2\sqrt{ x+1 } } } } $$
$$ = \displaystyle{ { 2 ( \infty)-1 \over { \sqrt{ {1 \over 1+1/{\infty}^2} } } + { 1 \over 2\sqrt{ \infty } } } } $$
$$ = \displaystyle{ \infty - 1 \over \sqrt{1 \over 1+0} + {1 \over \infty} } $$
$$ = \displaystyle{ \infty \over 1 + {0} } $$
$$ = \displaystyle{ \infty } $$
(The limit does not exist.)
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