I was trying to find $\lim_{z\rightarrow{0}} \frac{z^2}{\cos(z)-1}$ and the solution guide just told me to do $\lim_{z\rightarrow{0}} \frac{\cos(z)-1}{z^2} = -\frac{1}{2}$ and then take the reciprocal of this, which is $-2$ as my answer.

Can we actually do that? Does this actually obey the limit laws?

  • $\begingroup$ Yes, we can actually do that, provided that none of the limits are $0$. $\endgroup$
    – DHMO
    Apr 17, 2017 at 14:47
  • 1
    $\begingroup$ You can do even more: if $\lim f(z)=a, \lim g(z)=b$ and $b\neq 0$ then $\lim \frac{f(z)}{g(z)}=\frac{a}{b}$. $\endgroup$ Apr 17, 2017 at 14:50
  • $\begingroup$ @QuangHoang Can it work in another way, such that if $\lim \frac{f(z)}{g(z)} = \frac{a}{b}$, then $\lim f(z) = a, \lim g(z) = b$? $\endgroup$
    – WeiShan Ng
    Apr 17, 2017 at 15:03
  • $\begingroup$ No, think $f(z)=2z, g(z)=z$, then $\lim_{z\to 2} \frac{f(z)}{g(z)}=\frac 21$ but $\lim g(z)\ne 1$. $\endgroup$ Apr 17, 2017 at 15:05
  • $\begingroup$ This is a simple consequence of "algebra of limits" provided the limits under consideration are non-zero. $\endgroup$
    – Paramanand Singh
    Apr 18, 2017 at 9:30

2 Answers 2


This is the justification. Suppose $g$ is continuous at $L$, and suppose $\lim\limits_{x\to a}f(x)=L$. Then $$\lim\limits_{x\to a}g(f(x)) = g(\lim\limits_{x\to a}f(x)) = g(L)$$ The first equality is from the continuity of $g$ at $L$, and the second is the existence of the original limit.

One needs to be a little more careful in the cases $x\to\pm\infty$ since the domain of $g$ typically doesn't include $\infty$.

In your case, $g(x)=1/x$, and your statement can be rearranged to be $$g(\lim f(x)) = \lim g(f(x))$$ by inverting both sides.

  • $\begingroup$ I get it now, thanks a lot! $\endgroup$
    – WeiShan Ng
    Apr 17, 2017 at 15:07

Here's another justification of it, just for fun.

In general, $$\lim_{x\to a} \frac{F(x)}{G(x)} = \frac{\displaystyle\lim_{x\to a} F(x)}{\displaystyle\lim_{x\to a} G(x)}$$ as long as both limits exist and $\displaystyle\lim_{x\to a} G(x) \ne 0$.

Applying that to your situation, we have:

$$ \lim_{x \to a} \frac1{f(x)} = \frac{\displaystyle\lim_{x\to a} 1}{\displaystyle\lim_{x\to a} f(x)} = \frac1{\displaystyle\lim_{x\to a} f(x)}$$


$$\frac1{\displaystyle\lim_{x\to a}\frac1{f(x)}} = \frac1{\left(\frac{\displaystyle\lim_{x\to a} 1}{\displaystyle\lim_{x\to a} f(x)}\right)} = \frac1{\left(\dfrac1{\displaystyle\lim_{x\to a} f(x)}\right)} = \lim_{x\to a} f(x)$$


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