babyagelesszj

2022-07-09

We have the following limits:
$\underset{x\to \frac{\pi }{4}}{lim}\frac{1-{\mathrm{tan}}^{2}x}{\mathrm{cos}2x}$ and $\underset{\theta \to 1}{lim}=\frac{\theta -1+\mathrm{sin}\left({\theta }^{2}-1\right)}{{\theta }^{2}-1}$
Using L'Hôpital I obtained that the second limit is $\theta =\frac{3}{2}$. For the first limit I was able to rewrite it to $\underset{x\to \frac{\pi }{4}}{lim}\left(1-{\mathrm{tan}}^{2}\left(x\right)\right)\mathrm{sec}\left(2x\right)$, but I don't really know what to do with that.

Marisol Morton

Expert

For the first limit, some elementary trigonometry (a duplication formula) will do:
$\frac{1-{\mathrm{tan}}^{2}x}{cos2x}=\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x\mathrm{cos}2x}=\frac{\mathrm{cos}2x}{{\mathrm{cos}}^{2}x\mathrm{cos}2x}=\frac{1}{{\mathrm{cos}}^{2}x}\underset{x\to \frac{\pi }{4}}{\overset{}{\to }}2.$
For the second limit, use asymptotic analysis:
$\frac{\theta -1+\mathrm{sin}\left({\theta }^{2}-1\right)}{{\theta }^{2}-1}=\frac{\theta -1+{\theta }^{2}-1+o\left({\theta }^{2}-1\right)}{{\theta }^{2}-1}=\frac{1}{\theta +1}+1+o\left(1\right)\underset{\theta \to 1}{\overset{}{\to }}\frac{1}{2}+1.$

Jonathan Miles

Expert

use the $\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=1$
$\frac{\theta -1+\mathrm{sin}\left({\theta }^{2}-1\right)}{{\theta }^{2}-1}=\frac{\theta -1}{{\theta }^{2}-1}+\frac{\mathrm{sin}\left({\theta }^{2}-1\right)}{{\theta }^{2}-1}$
$=\frac{1}{\theta +1}+\frac{\mathrm{sin}\left({\theta }^{2}-1\right)}{{\theta }^{2}-1}$
$\underset{x\to 1}{lim}\left(\frac{1}{\theta +1}+\frac{\mathrm{sin}\left({\theta }^{2}-1\right)}{{\theta }^{2}-1}\right)=1/2+1=3/2$

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