# Clarification on proof of $\lim_{n\to +\infty}\left(\frac{f(a+\frac{1}{n})}{f(a)}\right)^{\frac{1}{n}}\to 1$

Let $$f:\mathbb{R}\to \mathbb{R}$$ be differentiable at $$a\in \mathbb{R}$$ such that $$f(a)>0$$. Evaluate: $$\lim_{n\to +\infty}\left(\frac{f(a+\frac{1}{n})}{f(a)}\right)^{\frac{1}{n}}.$$

Attempt. A proof would go like: $$\begin{eqnarray} \left(\frac{f(a+\frac{1}{n})}{f(a)}\right)^{\frac{1}{n}}&=&\exp\left\{\ln\left(\frac{f(a+\frac{1}{n})}{f(a)}\right)^{\frac{1}{n}}\right\}= \exp\left\{\frac{\ln f(a+\frac{1}{n})-\ln f(a)}{n}\right\}\nonumber\\ &=&\exp\left\{\frac{1}{n^2}\,\frac{\ln f(a+\frac{1}{n})-\ln f(a)}{\frac{1}{n}}\right\}\nonumber\\ &\to & \exp\left\{0\cdot \frac{f'(a)}{f(a)}\right\}=1,~n\to +\infty,\nonumber \end{eqnarray}$$ where we use the definition of derivative and the chain rule. So far, so good.

My question is:

one claims that since $$f$$ is differentible at $$a,$$ then $$f$$ is continuous at $$a$$ and so the limit becomes $$1^0=1$$, according to the algebra of sequential limits.

Is such an approach also correct? ($$1^0$$ is not an indeterminate form).

Thanks for the help.

• Since $f$ is continuous at $x=a$ (follows from the differentiability at $a$), $\lim_{x \to a} e^{f(x)} = e^{\lim_{x \to a} f(x)}$ Jan 22, 2020 at 23:27

Want $$\lim_{n\to +\infty}\left(\dfrac{f(a+\frac{1}{n})}{f(a)}\right)^{\frac{1}{n}}.$$

I would write, since $$f(a+h) \approx f(a)+hf'(a)$$,

$$\begin{array}\\ \dfrac{f(a+\frac{1}{n})}{f(a)} &\approx \dfrac{f(a)+\frac1{n}f'(a)}{f(a)}\\ &= 1+ \dfrac{f'(a)}{nf(a)}\\ \text{so}\\ \left(\dfrac{f(a+\frac{1}{n})}{f(a)}\right)^{\frac{1}{n}} &\approx \left(1+ \dfrac{f'(a)}{nf(a)}\right)^{\frac{1}{n}}\\ &\approx 1+ \dfrac{f'(a)}{n^2f(a)}\\ &\to 1\\ \end{array}$$

Question: Is that exponent $$\dfrac1{n}$$ or $$n$$? If it is $$n$$, we have a much more interesting result:

$$\begin{array}\\ \left(\dfrac{f(a+\frac{1}{n})}{f(a)}\right)^{n} &\approx \left(1+ \dfrac{f'(a)}{nf(a)}\right)^{n}\\ &\approx \left(\left(1+ \dfrac{f'(a)}{nf(a)}\right)^{\dfrac{nf(a)}{f'(a)}}\right)^{\dfrac{f'(a)}{f(a)}}\\ &\to e^{\dfrac{f'(a)}{f(a)}}\\ \end{array}$$

Your argument at the end is correct. The expression under parentheses tends to $$1$$ and the exponent $$1/n$$ tends to $$0$$ and hence the desired limit is $$1^0=1$$.

Note the following rule:

If $$f(x) \to a>0$$ and $$g(x) \to b$$ as $$x\to c$$ then $$\{f(x) \} ^{g(x)} \to a^b$$ as $$x\to c$$.

Try to prove the above using exponential and logarithmic functions.

For current question you only need that $$f$$ is continuous as $$a$$ and $$f(a) \neq 0$$.