I'm trying to evaluate the following complex limit:

$$\lim_{z \rightarrow -1} \left|\frac{\sqrt{z^2-1}}{z+1}\right|.$$

I notice that the limit is of the indeterminate form $0/0$ which prompts me to use l'Hopital's rule to get infinity. However, my instincts has told me that this is not quite a valid use for l'Hopital's rule, although I'm not clear on why I feel this way. As such, I have tried another method of evaluating this limit by rearranging it to look like the following expression:

$$\lim_{z \rightarrow -1} \left|\sqrt{1 - \frac{2}{z + 1}}\right|.$$

Again this evaluates to infinity and seems to be a little more acceptable than the previous method. As such, I would like to verify whether any of these methods are correct, and if not, how would one evaluate this limit without resorting to the use of Wolfram Alpha say.

Thanks for the help!

  • 1
    $\begingroup$ Your methods are correct. $\endgroup$ Nov 16, 2019 at 5:43
  • $\begingroup$ Hopital rule gives 1 as the limit. $\endgroup$
    – sirous
    Nov 16, 2019 at 7:27
  • $\begingroup$ I'm afraid I don't see how l'Hopital's rule gives 1 as the limit... $\endgroup$ Nov 16, 2019 at 9:12

1 Answer 1


Both methods are correct. Using L'Hôpital's rule and continuity of the absolute value function

$$\lim_{z \rightarrow -1^+} \left|\frac{\sqrt{z^2-1}}{z+1}\right|=\left|\lim_{z \rightarrow -1^+}\frac{z}{\sqrt{z^2-1}}\right|=\left|\lim_{z \rightarrow -1^+}\frac{1/z}{\sqrt{1-\frac{1}{z^2}}}\right|=\lim_{z \rightarrow -1^+}\frac{1}{\sqrt{\left|1-\frac{1}{z^2}\right|}}=\infty$$ $$\lim_{z \rightarrow -1^-} \left|\frac{\sqrt{z^2-1}}{z+1}\right|=\left|\lim_{z \rightarrow -1^-}\frac{z}{\sqrt{z^2-1}}\right|=\left|\lim_{z \rightarrow -1^-}\frac{1/z}{\sqrt{1-\frac{1}{z^2}}}\right|=\lim_{z \rightarrow -1^-}\frac{1}{\sqrt{\left|1-\frac{1}{z^2}\right|}}=\infty$$


$$\left|\lim_{z \rightarrow -1^+}\sqrt{1 - \frac{2}{z + 1}}\right|=\lim_{z \rightarrow -1^+}\sqrt{\left|1 - \frac{2}{z + 1}\right|}=\infty$$ $$\left|\lim_{z \rightarrow -1^-}\sqrt{1 - \frac{2}{z + 1}}\right|=\lim_{z \rightarrow -1^-}\sqrt{\left|1 - \frac{2}{z + 1}\right|}=\infty$$

  • $\begingroup$ Thanks for the verification! $\endgroup$ Nov 17, 2019 at 2:13
  • $\begingroup$ @MathsMatador: Also observe that L'Hôpital's rule is valid here since $\sqrt{z^2-1}$ and ${z+1}$ are holomorphic in a neighborhood of $-1$. $\endgroup$
    – Axion004
    Nov 17, 2019 at 3:40

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