Show that if $f : (a,\infty)\to \mathbb R$ is such that $\lim_{x\to \infty}xf(x)=L$ where $ L \in \mathbb R$, then $\lim_{x\to \infty}f(x)=0$

Let $\epsilon>0$,$\exists K(\epsilon)>0$: $\forall x:|x|>K(\epsilon)\implies \left|\left|xf(x)\right|-\left|L\right|\right| \leq|xf(x)-L|<\epsilon$

$\epsilon>0$,$\exists K(\epsilon)>0$: $\forall x:|x|>K(\epsilon)\implies \frac{|L|-\epsilon}{|x|}<|f(x)|<\frac{|L|+\epsilon}{|x|}$

Choose $M=\max\left\{\frac{|L|+\epsilon}{\epsilon}, \, K(\epsilon)\right\}\implies \forall x>M ,|f(x)|<\epsilon$

Am I correct? Can I choose $M$ like this?

  • 1
    $\begingroup$ Please see the minor corrections that I made to your solution. I am not sure what the problem tells you about $a$ but if $a$ is finite then technically we don't have to worry about $|x|$ since we could then interpret $x \to \infty$ to mean $x \to +\infty$. $\endgroup$
    – M A Pelto
    May 19, 2018 at 1:20
  • $\begingroup$ Thank you@MattAPelto $\endgroup$
    – user464147
    May 19, 2018 at 15:59

2 Answers 2


Let $x\to \infty$ in


and apply the squeeze theorem.

  • 1
    $\begingroup$ Can you please check whether my arguments are correct or not? $\endgroup$
    – user464147
    May 18, 2018 at 15:43
  • $\begingroup$ I think your arguments can possibly be made correct, but you need to pay attention to the signs of $\frac{L-\epsilon}{x}$ and $\frac{L+\epsilon}{x}$. $\endgroup$ May 18, 2018 at 15:45
  • 2
    $\begingroup$ Notice that even if $x>0$, $x>M$ does not imply $\frac1M > \frac 1x$ -- it fails when $M<0$ and does not make sense for $M=0$. $\endgroup$ May 18, 2018 at 15:47
  • $\begingroup$ I think I should consider two cases $x>0$ and $x<0$. Right? $\endgroup$
    – user464147
    May 18, 2018 at 15:50
  • $\begingroup$ You don't need to consider the case $x<0$ because you're concerned about the behavior as $x\to+\infty$. However, you need to consider the signs of $L-\epsilon$ and $L+\epsilon$. $\endgroup$ May 18, 2018 at 15:54

1) Let $\epsilon >0$ be given.

Need to show:

There is a $K$, real, positive, such that

$x>K$ implies $|f(x)| < \epsilon$.

2) Given:

Let $\varepsilon >0$ be given.

There is a $M$, real, positive, such that

$x \gt M$ implies $|xf(x)-L| < \varepsilon.$

3) There is a $N$, real, positive,

such $x >N$ implies

$|1/x| \lt \min (1/|L|, 1/\varepsilon) \epsilon/2.$

Let $K= \max(M,N)$.

Then for $x \gt K$:

$|f(x)| = |1/x||xf(x)|=$

$|1/x| |xf(x)-L+L| \le$

$|1/x||xf(x)-L| +|1/x||L| \lt$

$|1/x|\varepsilon +|1/x||L| \lt \epsilon/2 +\epsilon/2 =\epsilon$.


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