# Find the limit lim_{xrightarrow0}frac{xsin x}{2-2cos x}

Tazmin Horton 2020-10-28 Answered
Find the limit
$\underset{x\to 0}{lim}\frac{x\mathrm{sin}x}{2-2\mathrm{cos}x}$
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To find the limit,
$\underset{x\to 0}{lim}\frac{x\mathrm{sin}x}{2-2\mathrm{cos}x}$
Let $l=\underset{x\to 0}{lim}\frac{x\mathrm{sin}x}{2-2\mathrm{cos}x}$
$⇒l=\underset{x\to 0}{lim}\frac{x\mathrm{sin}x}{2\left(1-\mathrm{cos}x\right)}$

$⇒l=\underset{x\to 0}{lim}\frac{x\mathrm{cos}\frac{x}{2}}{2\mathrm{sin}\frac{x}{2}}$
$⇒l=\underset{x\to 0}{lim}\frac{\frac{x}{2}}{\mathrm{sin}\frac{x}{2}}\underset{x\to 0}{lim}\mathrm{cos}\frac{x}{2}$

$⇒l=\underset{y\to 0}{lim}\frac{1}{\mathrm{cos}y}$ (Applying LHospitals Rule)
$⇒l=\frac{1}{\mathrm{cos}0}$
$⇒l=1$
Therefore, $\underset{x\to 0}{lim}\frac{x\mathrm{sin}x}{2-2\mathrm{cos}x}=1$