Find: $\lim_{n\longrightarrow\infty}\left(\frac{f(a+\frac{1}{n})}{f(a)}\right)^{n}$ as $f$ is differentiable on $x=a$, and $f'(a)=2, f(a)=5$

Question

Let $f$ be defined in the environment of the point $a$. Suppose $f$ is differentiable on $x=a$, and $f'(a)=2,\ f(a)=5$

Find: $$\lim_{n\longrightarrow\infty}\left(\frac{f(a+\frac{1}{n})}{f(a)}\right)^{n}$$

I've tried to use the fact that according to the Differential of a function.

let $a=x_{0}$. If $f$ is differentiable in $x=x_{0}$ than, the line: $$l(x) = f(x_{0}) + f'(x_{0})(x-x_{0})$$ exists: $$\lim_{x\longrightarrow x_{0}}\frac{f(x)-l(x)}{x-x_{0}}=0$$

and then $$\lim_{n\longrightarrow\infty}\left(\frac{f(a+\frac{1}{n})}{f(a)}\right)^{n}=\lim_{n\longrightarrow\infty}\left(\frac{f(a)+f(a+\frac{1}{n})-f(a)}{f(a)}\right)^{n}=l$$

but I didn't managed to get some progress from here.

Answer 1

Note that

$$\ln\left(\frac{f\left(a + \frac{1}{n}\right)}{f(a)}\right)^n = \frac{\ln\left(f\left(a + \frac{1}{n}\right)\right) - \ln(f(a))}{\frac{1}{n}}$$

As $n \to \infty$, this tends to the derivative of $\ln \circ f$ at $a$. Using chain rule, this limit comes to $$\frac{f'(a)}{f(a)} = \frac{2}{5}.$$

However, of course, the original limit is $e^{\frac{2}{5}}$.

Answer 2

$$\lim_{n\longrightarrow\infty}\left(\frac{f(a+\frac{1}{n})}{f(a)}\right)^{n}\\=\lim_{n\longrightarrow\infty}\left(\frac{f(a)+\frac{1}{n}f'(a)+...}{f(a)}\right)^{n}\\=\lim_{n\longrightarrow\infty}(1+(2/5)(1/n))^n=e^{2/5} $$