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.)