0
$\begingroup$

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?

$\endgroup$
13
  • 2
    $\begingroup$ That is incorrect, you can't conclude anything from $\infty\cdot 0$. The limit does exist and equals something nice. $\endgroup$ Commented May 21, 2021 at 20:05
  • $\begingroup$ @NinadMunshi I see. What is incorrect? $\endgroup$
    – mvfs314
    Commented May 21, 2021 at 20:06
  • $\begingroup$ Note that the function is differentiable at $x=0$. What is derivative of f at $0$? $\endgroup$
    – Koro
    Commented May 21, 2021 at 20:06
  • $\begingroup$ Hint: Are you familiar with L'hospital's rule? $\endgroup$ Commented May 21, 2021 at 20:07
  • 2
    $\begingroup$ 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 $$\lim_{n \rightarrow \infty} 1 = \lim_{n \rightarrow \infty} n \cdot \lim_{n \rightarrow \infty} \frac{1}{n} \text{.} $$ $\endgroup$ Commented May 21, 2021 at 20:12

2 Answers 2

4
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ I thought that would be understood in these kinds of contexts - where we leave out the $x_n$ which are zero. $\endgroup$
    – Yathi
    Commented May 22, 2021 at 14:18
2
$\begingroup$

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)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .