Answer
$a.\displaystyle \quad\lim_{n\rightarrow\infty}a_{n}=f'(0) \qquad $(proof given below)
$b.\quad 1$
$c.\quad 1$
$d.\quad 2$
Work Step by Step
$ a.\quad$
$\begin{aligned}
\displaystyle \lim_{n\rightarrow\infty}a_{n}&=\displaystyle \lim_{n\rightarrow\infty}n\cdot f(\frac{1}{n}) \quad\text{... substitute } h=\displaystyle \frac{1}{n} \\
&\text{ when } n\rightarrow\infty, \text{ then } h\rightarrow 0^{+}& & \\
&=\displaystyle \lim_{h\rightarrow 0^{+}}\frac{f(h)}{h} & & \\
& =\displaystyle \lim_{h\rightarrow 0^{+}}\frac{f(h)-0}{h}\quad \text{... use: } f(0)=0 & \\
& =\displaystyle \lim_{h\rightarrow 0^{+}}\frac{f(h)-f(0)}{h} \quad \text{...definition of the derivative }\\
&=f'(0) \end{aligned}$
$ b.\quad$
$f(x)=\tan^{-1}x$
$f'(x)=\displaystyle \frac{1}{1+x^{2}}$
$\displaystyle \lim_{n\rightarrow\infty}a_{n}=f'(0)=\frac{1}{1+(0^{2})}=1$
$ c.\quad$
$f(x)=e^{x}-1$
$f'(x)=e^{x}$
$\displaystyle \lim_{n\rightarrow\infty}a_{n}=f'(0)=e^{0}=1$
$ d.\quad$
$f(x)=\ln(1+2x)$
$f'(x)=\displaystyle \frac{1}{1+2x}\cdot 2=\frac{2}{1+2x}$
$\displaystyle \lim_{n\rightarrow\infty}a_{n}=f'(0)=\frac{2}{1+0}=2$
