# Evaluate the following limits. lim_{xrightarrow0}frac{tan7x}{sin x}

Evaluate the following limits.
$\underset{x\to 0}{lim}\frac{\mathrm{tan}7x}{\mathrm{sin}x}$
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broliY
Given Function,
$f\left(x\right)=\frac{\mathrm{tan}7x}{\mathrm{sin}x}$
We have to find,
$\underset{x\to 0}{lim}\frac{\mathrm{tan}7x}{\mathrm{sin}x}$
As we know,
If we get $\frac{0}{0}$ an Indeterminate form then,
$\underset{x\to 0}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to 0}{lim}\frac{{f}^{\prime }\left(x\right)}{{g}^{\prime }\left(x\right)}$
$\underset{x\to 0}{lim}\frac{\mathrm{tan}7x}{\mathrm{sin}x}$
We have an indeterminate form of type $\frac{0}{0}$, we can apply L'Hospital Rule,
$\underset{x\to 0}{lim}\frac{\mathrm{tan}7x}{\mathrm{sin}x}=\underset{x\to 0}{lim}\frac{\frac{d}{dx}\left(\mathrm{tan}7x\right)}{\frac{d}{dx}\left(\mathrm{sin}x\right)}$
$\underset{x\to 0}{lim}\frac{\mathrm{tan}7x}{\mathrm{sin}x}=\underset{x\to 0}{lim}\frac{7{\mathrm{tan}}^{2}7x+7}{\mathrm{cos}x}$
as we know,
${\mathrm{tan}}^{2}x={\mathrm{sec}}^{2}x-1$
$\underset{x\to 0}{lim}\frac{\mathrm{tan}7x}{\mathrm{sin}x}=\underset{x\to 0}{lim}\frac{7\left({\mathrm{sec}}^{2}7x-1\right)+7}{\mathrm{cos}x}$
$=\underset{x\to 0}{lim}\frac{7{\mathrm{sec}}^{2}7x-7+7}{\mathrm{cos}x}$
$=\underset{x\to 0}{lim}\frac{7}{\mathrm{cos}x×{\mathrm{cos}}^{2}7x}$
put x=0
$\underset{x\to 0}{lim}\frac{\mathrm{tan}7x}{\mathrm{sin}x}=\frac{7}{1}=7$
Result: $\underset{x\to 0}{lim}\frac{\mathrm{tan}7x}{\mathrm{sin}x}=7$
Jeffrey Jordon