2. Evaluate the limit if it exists or explain why it does not exist. (Do not use L'hospital rules!) (a) $\lim _{x ightarrow-1} \frac{2-\sqrt{f(x)}}{x+1}$ if $f(-1)=4, f(x)>0$ for all $x$ and $f^{\prime} (-1)=\pi$. (b) $\lim _{x ightarrow \infty} \frac{\ln \left(1+\sin ^{2} x ight)}{\arctan X+e^{x}}$ (c) $\lim _{x ightarrow 0} \frac{ \sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}}$ (d) $\lim _{x ightarrow \infty} \frac{x^{3}+\sqrt{x^{6}+2 x^{5}+1}}{x^{3}+2 X^{2}+X+1}$ (e) $\lim _{x ightarrow \infty} \frac{\cos (x \sin x)}{x^{2}}$ (f) $\lim _{x ightarrow 0^{+}} \frac{(\tan \sqrt{x}) \sqrt{2 x+1}}{x}$ (g) $\lim _{x ightarrow \infty} \sqrt{x+\sqrt{x}}-\sqrt{x}$ (h) $\lim _{x ightarrow \infty}\left[\In \left(x^{2}+1 ight)-\In \left(x^{2}-1 ight) ight]$ CS. JG. 118