# What is the value of lim_(x->0) (1-cos(mx))/(xsinx)?

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Mar 8, 2018

${m}^{2} / 2$.

#### Explanation:

$\text{The Reqd. Lim.=} {\lim}_{x \to 0} \frac{1 - \cos m x}{x \sin x}$,

$= {\lim}_{x \to 0} \frac{2 {\sin}^{2} \left(\frac{m x}{2}\right)}{x \sin x}$,

$= \lim 2 {\left\{\sin \frac{\frac{m x}{2}}{\frac{m x}{2}} \cdot \left(\frac{m x}{2}\right)\right\}}^{2} \div \left\{x \left(\frac{\sin x}{x} \cdot x\right)\right\}$,

$= \lim 2 {\left\{\sin \frac{\frac{m x}{2}}{\frac{m x}{2}}\right\}}^{2} \cdot \frac{{m}^{2} {x}^{2}}{4} \div \left\{{x}^{2} \left(\sin \frac{x}{x}\right)\right\}$,

$= \left(2 \cdot {m}^{2} / 4\right) \cdot {\lim}_{\frac{m x}{2} \to 0} {\left\{\sin \frac{\frac{m x}{2}}{\frac{m x}{2}}\right\}}^{2} \div {\lim}_{x \to 0} \left\{\left(\sin \frac{x}{x}\right)\right\}$,

$= {m}^{2} / 2 \cdot {\left(1\right)}^{2} \div 1$.

$\Rightarrow \text{The Reqd. Lim.=} {m}^{2} / 2$.

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