# Evaluating trigonometric limit $$\displaystyle{{\csc}^{{2}}{\left({2}{x}\right)}}-{\frac{{{1}}}{{{4}{x}^{{2}}}}}$$

Evaluating trigonometric limit ${\mathrm{csc}}^{2}\left(2x\right)-\frac{1}{4{x}^{2}}$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

arrebusnaavbz
Based on DeepSea's hint, I've managed to solve by myself.
$\underset{x\to 0}{lim}\left[{\mathrm{csc}}^{2}\left(2x\right)-\frac{1}{4{x}^{2}}\right]$
$=\underset{x\to 0}{lim}\left[\frac{1}{{\mathrm{sin}}^{2}\left(2x\right)}-\frac{1}{4{x}^{2}}\right]$
$=\underset{x\to 0}{lim}\left[\frac{4{x}^{2}-{\mathrm{sin}}^{2}\left(2x\right)}{4{x}^{2}\left({\mathrm{sin}}^{2}\left(2x\right)\right)}\right]$
$=\underset{x\to 0}{lim}\left[\frac{4{x}^{2}-{\mathrm{sin}}^{2}\left(2x\right)}{4{x}^{2}×4{x}^{2}{\left(\frac{\mathrm{sin}\left(2x\right)}{2x}\right)}^{2}}\right]$
$=\underset{x\to 0}{lim}\left[\frac{4{x}^{2}-{\mathrm{sin}}^{2}\left(2x\right)}{4{x}^{2}×4{x}^{2}}\right]×\underset{x\to 0}{lim}\left[\frac{1}{{\left(\frac{\mathrm{sin}\left(2x\right)}{2x}\right)}^{2}}\right]$
$=\underset{x\to 0}{lim}\left[\frac{4{x}^{2}-{\mathrm{sin}}^{2}\left(2x\right)}{16{x}^{4}}\right]×1$
$=\underset{x\to 0}{lim}\left[\frac{8x-2\mathrm{sin}\left(4x\right)}{64{x}^{3}}\right]$
$=\underset{x\to 0}{lim}\left[\frac{8-8\mathrm{cos}\left(4x\right)}{192{x}^{2}}\right]$
$=\underset{x\to 0}{lim}\left[\frac{1-\mathrm{cos}\left(4x\right)}{24{x}^{2}}\right]$
$=\frac{1}{24}×\underset{x\to 0}{lim}\left[\frac{1-\mathrm{cos}\left(4x\right)}{{x}^{2}}\right]$
$=\frac{1}{24}×\frac{1}{2}×{4}^{2}$
$=\frac{1}{3}$
###### Not exactly what you’re looking for?
Lana Hamilton
hint: replace the denominator on the third line of your proof and apply L'hopitale rule three times to the expression: $\frac{4{x}^{2}-{\mathrm{sin}}^{2}\left(2x\right)}{16{x}^{4}}$.I did it and it works. Try it. Note that "replace" here means you write: ${\mathrm{sin}}^{2}\left(2x\right)=4{x}^{2}\cdot {\left(\frac{\mathrm{sin}\left(2x\right)}{2x}\right)}^{2}$ and the second factor approaches 1 when $x\to 0$