Evaluating $\displaystyle\lim_{n\to\infty}nf\left(\frac{1}{n}\right)$

Let $$f\colon\mathbb{R}\to\mathbb{R}$$ a differentiable function with $$f(0)=0$$. I have to evaluate the limit: $$\lim_{n\to\infty}nf\left(\frac{1}{n}\right)$$

Since the limit $$\lim_{n\to\infty}\frac{1}{n}=0$$ and $$f$$ is continuous due to its differentiability, then $$\lim_{n\to\infty}f\left(\frac{1}{n}\right)=f\left(\lim_{n\to\infty}\frac{1}{n}\right)=f(0)=0$$ so, the limit would be: $$\lim_{n\to\infty}nf\left(\frac{1}{n}\right)=\lim_{n\to\infty}n\cdot\lim_{n\to\infty}f\left(\frac{1}{n}\right)=\infty\cdot0$$ so may I conclude this limit doesn't exist?

• That is incorrect, you can't conclude anything from $\infty\cdot 0$. The limit does exist and equals something nice. Commented May 21, 2021 at 20:05
• @NinadMunshi I see. What is incorrect? Commented May 21, 2021 at 20:06
• Note that the function is differentiable at $x=0$. What is derivative of f at $0$?
– Koro
Commented May 21, 2021 at 20:06
• Hint: Are you familiar with L'hospital's rule? Commented May 21, 2021 at 20:07
• You should check your product rule for limits again: the rewrite $\lim a b = \lim a \cdot \lim b$ requires that the limits on the right both exist. To convince yourself that this is required, consider $$\lim_{n \rightarrow \infty} 1 = \lim_{n \rightarrow \infty} n \cdot \lim_{n \rightarrow \infty} \frac{1}{n} \text{.}$$ Commented May 21, 2021 at 20:12

$$\frac{1}{n}$$ is a sequence converging to 0. $$f$$ is differentiable at 0. Therefore, for any sequence $$x_n$$ converging to 0, $$\frac{f(x_n) - f(0)}{x_n}$$, must converge to $$f^{\prime}(0)$$. Hence the limit exists and its value is $$f^{\prime}(0)$$.
• I thought that would be understood in these kinds of contexts - where we leave out the $x_n$ which are zero. Commented May 22, 2021 at 14:18
Since $$\displaystyle f'( 0)$$ exists, it follows that the limit $$\begin{equation*} \lim _{x\rightarrow 0}\frac{f( x) -f( 0)}{x-0} =\lim _{x\rightarrow 0}\frac{f( x)}{x} =f'( 0) \end{equation*}$$ Hence the limit $$\begin{equation*} \lim _{x\rightarrow 0}\frac{f( x)}{x} =f'( 0) \end{equation*}$$ That is for every sequence $$\displaystyle x_{n}\rightarrow 0$$ (such that $$\displaystyle x_{n} \neq 0$$ for any $$\displaystyle n$$), $$\displaystyle \frac{f( x_{n})}{x_{n}}\rightarrow f'( 0)$$
In particular, $$\displaystyle \frac{1}{n}\rightarrow 0\Longrightarrow \displaystyle \frac{f\left(\frac{1}{n}\right)}{\frac{1}{n}}\rightarrow f'( 0)$$.