Problems with finding limit The function $f(x)$ has a derivative at the point $a$ and  $f(a) > 0$. I need to find the limit as n $\to + \infty$ of $$\left(\frac{f(a + \frac1n)}{f(a)} \right)^n$$ Substitution method?
 A: You care about the limit
$$
\lim_{n\to \infty}\left(\frac{f(a+1/n)}{f(a)}\right )^n
$$
which is the same as 
$$
\lim_{n\to \infty}
\exp\left(n\log\left(\frac{f(a+1/n)}{f(a)}\right )\right)\\
=\lim_{n\to \infty}\exp\left(\frac{\log\left(\frac{f(a+1/n)}{f(a)}\right )}{1/n}\right)
$$
Where we note that the expression in the exponential is the derivative of 
the function 
$$
g(x)=\log(f(x))
$$
at $a$ which by your assumption, exists. So the limit is in fact
$$
\exp\left(\frac{f'(a)}{f(a)}\right)
$$
A: $$\left(\frac{f(a + \frac1n)}{f(a)} \right)^n=e^{n \log \left(\frac{f(a + \frac1n)}{f(a)} \right)}$$
$$n \log \left(\frac{f(a + \frac1n)}{f(a)} \right)=n[\log {f(a + \frac1n)}-\log f(a)])=\frac{[\log {f(a + \frac1n)}-\log f(a)]}{\frac1n}=\frac{\log {f(a + h)}-\log f(a)}{h}\to \frac{f'(a)}{f(a)} $$
thus
$$\left(\frac{f(a + \frac1n)}{f(a)} \right)^n \to e^{\frac{f'(a)}{f(a)} }$$
A: We know there exists $\xi_n\in(a,a+1/n)$ such that $f(a+1/n)-f(a)=\frac{1}{n}f'(\xi_n)$. Now as $n\rightarrow\infty$, $f'(\xi_n)\rightarrow f'(a)$. So the limit becomes $$\lim_{n\rightarrow\infty}\left(\frac{f(a+1/n)}{f(a)}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{f'(\xi_n)}{nf(a)}\right)^n=e^{\frac{f'(a)}{f(a)}}$$
A: By definition of derivative we get 
$$ \lim_{n\to \infty}\left(\frac{f(a + \frac1n)}{f(a)} \right)^n= \lim_{n\to \infty}
\exp\left[n\log\left(\frac{f(a+1/n)}{f(a)}\right )\right]
=\lim_{h\to 0}\exp\left[\frac{\log\left(f(a+h\right ) -\log f(a)}{h}\right] =\exp\left(\frac{f'(a)}{f(a)}\right)$$
