babyagelesszj

Answered

2022-07-09

We have the following limits:

$\underset{x\to \frac{\pi}{4}}{lim}{\displaystyle \frac{1-{\mathrm{tan}}^{2}x}{\mathrm{cos}2x}}$ and $\underset{\theta \to 1}{lim}={\displaystyle \frac{\theta -1+\mathrm{sin}({\theta}^{2}-1)}{{\theta}^{2}-1}}$

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

$\underset{x\to \frac{\pi}{4}}{lim}{\displaystyle \frac{1-{\mathrm{tan}}^{2}x}{\mathrm{cos}2x}}$ and $\underset{\theta \to 1}{lim}={\displaystyle \frac{\theta -1+\mathrm{sin}({\theta}^{2}-1)}{{\theta}^{2}-1}}$

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

Answer & Explanation

Marisol Morton

Expert

2022-07-10Added 13 answers

For the first limit, some elementary trigonometry (a duplication formula) will do:

$\frac{1-{\mathrm{tan}}^{2}x}{cos2x}=\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x\mathrm{cos}2x}=\frac{\mathrm{cos}2x}{{\mathrm{cos}}^{2}x\mathrm{cos}2x}=\frac{1}{{\mathrm{cos}}^{2}x}\underset{x\to {\scriptstyle \frac{\pi}{4}}}{\overset{}{\to}}2.$

For the second limit, use asymptotic analysis:

$\frac{\theta -1+\mathrm{sin}({\theta}^{2}-1)}{{\theta}^{2}-1}=\frac{\theta -1+{\theta}^{2}-1+o({\theta}^{2}-1)}{{\theta}^{2}-1}=\frac{1}{\theta +1}+1+o(1)\underset{\theta \to 1}{\overset{}{\to}}\frac{1}{2}+1.$

$\frac{1-{\mathrm{tan}}^{2}x}{cos2x}=\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x\mathrm{cos}2x}=\frac{\mathrm{cos}2x}{{\mathrm{cos}}^{2}x\mathrm{cos}2x}=\frac{1}{{\mathrm{cos}}^{2}x}\underset{x\to {\scriptstyle \frac{\pi}{4}}}{\overset{}{\to}}2.$

For the second limit, use asymptotic analysis:

$\frac{\theta -1+\mathrm{sin}({\theta}^{2}-1)}{{\theta}^{2}-1}=\frac{\theta -1+{\theta}^{2}-1+o({\theta}^{2}-1)}{{\theta}^{2}-1}=\frac{1}{\theta +1}+1+o(1)\underset{\theta \to 1}{\overset{}{\to}}\frac{1}{2}+1.$

Jonathan Miles

Expert

2022-07-11Added 3 answers

use the $\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=1$

$\frac{\theta -1+\mathrm{sin}({\theta}^{2}-1)}{{\theta}^{2}-1}}=\frac{\theta -1}{{\theta}^{2}-1}+\frac{\mathrm{sin}({\theta}^{2}-1)}{{\theta}^{2}-1$

$=\frac{1}{\theta +1}+\frac{\mathrm{sin}({\theta}^{2}-1)}{{\theta}^{2}-1}$

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

$\frac{\theta -1+\mathrm{sin}({\theta}^{2}-1)}{{\theta}^{2}-1}}=\frac{\theta -1}{{\theta}^{2}-1}+\frac{\mathrm{sin}({\theta}^{2}-1)}{{\theta}^{2}-1$

$=\frac{1}{\theta +1}+\frac{\mathrm{sin}({\theta}^{2}-1)}{{\theta}^{2}-1}$

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

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