# Regarding limits: $\lim\limits_{n\to \infty}\left(\frac{f\left(x+\frac{1}{n}\right)}{f(x)}\right)^n$

If $f$ is positive and differentiable in $(0,\infty)$, then I want to find the following limit.

$\lim\limits_{n\to \infty}\left(\dfrac{f\left(x+\dfrac{1}{n}\right)}{f(x)}\right)^n$.

I have done as follows: $\lim\limits_{n\to \infty}\left(\dfrac{f\left(x+\dfrac{1}{n}\right)}{f(x)}\right)^n=\left(\lim\limits_{n\to \infty}\dfrac{f\left(x+\dfrac{1}{n}\right)}{f(x)}\right)^n=1$ as $f$ is continuous at $x=\dfrac{1}{n}$. Am I right? I doubt. Please help!

• Can you explain your doubt in more detail? – Eric Towers Mar 9 '14 at 3:11
• Hint: take the function $f(x)=x$, and let's see what happens at $x=1$. We then have $lim_{n \to \infty}(1+1/n)^n$, which as you probably know does not equal $1$. – Théophile Mar 9 '14 at 3:13
• As $n\to \infty$, can I write $\lim\limits_{n\to \infty}x^n=(\lim\limits_{n\to \infty}x)^n$? – Anupam Mar 9 '14 at 3:13
• If you write that, Anupam, what will you mean by $\lim_{n\to\infty}x$? – Gerry Myerson Mar 9 '14 at 4:00
• Also see this question (math.stackexchange.com/q/618169/72031). – Paramanand Singh Mar 9 '14 at 6:52

$$\log\left[\left(\dfrac{f\left(x+\dfrac{1}{n}\right)}{f(x)}\right)^n \right]= n\log \dfrac{f(x) + \frac 1n f'(x)+\epsilon(\frac 1n)\frac 1n}{f(x) } = n\log \left[1+ \frac 1n\dfrac{ f'(x)}{f(x)}+\epsilon\left(\frac 1n\right)\frac 1n\right]$$ with $\lim_0\epsilon = 0$. $$\sim n \frac 1n\dfrac{ f'(x)}{f(x)} = \dfrac{ f'(x)}{f(x)}$$so the limit is $$\exp\dfrac{ f'(x)}{f(x)}$$

• It seems you are using $f(x+n^{-1})=f(x)+n^{-1}f'(x)$, which is only true as an approximation. – Gerry Myerson Mar 9 '14 at 4:29
• You have also forgotten to exponentiate at the end. – Gerry Myerson Mar 9 '14 at 4:33
• right, thanks. {} – mookid Mar 9 '14 at 4:34

Hint: take logarithms, and compare what you get to the difference-quotient definition of the derivative.

If $f$ is differentiable, we can write $f(x+{1 \over n}) = f(x) + {1 \over n} f'(x) + r({1 \over n})$, where $\lim_n {r({1 \over n}) \over {1 \over n}} = 0$.

In particular, for any $\epsilon>0$ we can find some $N$ such that if $n \ge N$ then $-\epsilon < {r({1 \over n}) \over {1 \over n}} < \epsilon$, and so ${1 \over n} (f'(x) - \epsilon) < {1 \over n} (f'(x) + {r({1 \over n}) \over {1 \over n}}) <{1 \over n} (f'(x) + \epsilon)$.

Hence we have $\left( 1+ {1 \over n}({ f'(x) - \epsilon \over f(x)}) \right)^n < \left( { f(x+{1 \over n}) \over f(x)} \right)^n \le \left( 1+ {1 \over n}({ f'(x) + \epsilon \over f(x)}) \right)^n$, and taking limits gives $e^{{ f'(x) - \epsilon \over f(x)}} \le \liminf_n \left( { f(x+{1 \over n}) \over f(x)} \right)^n \le \limsup_n \left( { f(x+{1 \over n}) \over f(x)} \right)^n \le e^{{ f'(x) + \epsilon \over f(x)} }$. Letting $\epsilon \downarrow 0$ yields $\lim_n \left( { f(x+{1 \over n}) \over f(x)} \right)^n = e^{{ f'(x) \over f(x)}}$.