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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.

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

2 Answers 2

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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} $

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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 $f(a) \neq 0$.

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