lugreget9

2021-12-30

$\underset{x\to \frac{\pi }{4}}{lim}\frac{\left({\mathrm{csc}}^{2}x-2{\mathrm{tan}}^{2}x\right)}{\left(\mathrm{cot}x-1\right)}$ without L'Hôpitals Rule.

Jenny Bolton

Hint:
$\frac{\left({\mathrm{csc}}^{2}x-2{\mathrm{tan}}^{2}x\right)}{\left(\mathrm{cot}x-1\right)}=\frac{{\mathrm{csc}}^{2}-2-2{\mathrm{tan}}^{2}x+2}{\mathrm{cot}x-1}$
$=\frac{{\mathrm{csc}}^{2}x-2}{\mathrm{cot}x-1}+2\mathrm{tan}x\frac{1-{\mathrm{tan}}^{2}x}{1-\mathrm{tan}x}$
$=\frac{{\mathrm{cot}}^{2}x-1}{\mathrm{cot}x-1}+2\mathrm{tan}x\frac{1-{\mathrm{tan}}^{2}x}{1-\mathrm{tan}x}$
$=\mathrm{cot}x+1+2\mathrm{tan}x\left(1+\mathrm{tan}x\right)$
It is easier to take the limit now

Serita Dewitt

This limit can actually be calculated quite directly. You only need $\mathrm{tan}\frac{\pi }{4}=1$
Rewrite the expression as follows and set $t=\mathrm{tan}x$ and consider $t\to 1$:
$\frac{\left({\mathrm{csc}}^{2}x-2{\mathrm{tan}}^{2}x\right)}{\left(\mathrm{cot}x-1\right)}=\frac{\frac{{\mathrm{cos}}^{2}x}{{\mathrm{sin}}^{2}x}+1-2{\mathrm{tan}}^{2}x}{\frac{1}{\mathrm{tan}x}-1}$
$\stackrel{t=\mathrm{tan}x}{=}\frac{\frac{1}{{t}^{2}}+1-2{t}^{2}}{\frac{1}{t}-1}$
$=\frac{1+{t}^{2}-2{t}^{4}}{t\left(1-t\right)}$
$=\frac{\left(1-t\right)\left(1+t\right)\left(1+2{t}^{2}\right)}{t\left(1-t\right)}$
$=\frac{\left(1+t\right)\left(1+2{t}^{2}\right)}{t}\stackrel{t\to 1}{\to }6$

Vasquez

I use (at the end)
$\begin{array}{}\mathrm{cos}x-\mathrm{sin}x=\sqrt{2}\left(\left(\frac{1}{\sqrt{2}}\right)\mathrm{cos}x-\left(\frac{1}{\sqrt{2}}\right)\mathrm{sin}x\right)\\ =\sqrt{2}\left(\mathrm{sin}\left(\frac{\pi }{4}\right)\mathrm{cos}x-\mathrm{cos}\left(\frac{\pi }{4}\right)\mathrm{sin}x\right)\\ =\sqrt{2}\mathrm{sin}\left(\frac{\pi }{4}-x\right)\\ v=\underset{x\to \frac{\pi }{4}}{lim}\frac{{\mathrm{csc}}^{2}x-2{\mathrm{tan}}^{2}x}{\mathrm{cot}x-1}\\ =\underset{x\to \frac{\pi }{4}}{lim}\frac{\frac{1}{{\mathrm{sin}}^{2}x}-2\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}}{\frac{\mathrm{cos}x}{}}\end{array}$

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