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?
Best Answer
$\frac{1}{n}$ 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 $f^{\prime}(0)$.