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$\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=$
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Verified Answer
The correct answer is:
$\frac{m^2}{n^2}$
$\begin{gathered}
\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=\lim _{x \rightarrow 0}\left\{\frac{2 \sin ^2 \frac{m x}{2}}{2 \sin ^2 \frac{n x}{2}}\right\} \\
=\lim _{x \rightarrow 0}\left[\left\{\frac{\sin \frac{m x}{2}}{\frac{m x}{2}}\right\}^2 \frac{m^2 x^2}{4} \cdot \frac{1}{\left\{\frac{\sin \frac{n x}{2}}{\frac{n x}{2}}\right\}^2} \cdot \frac{4}{n^2 x^2}\right] \\
=\frac{m^2}{n^2} \times 1=\frac{m^2}{n^2} .
\end{gathered}$
Aliter : Apply L-Hospital's rule,
$\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=\lim _{x \rightarrow 0} \frac{m \sin m x}{n \sin n x}=\lim _{x \rightarrow 0} \frac{m^2 \cos m x}{n^2 \cos n x}=\frac{m^2}{n^2} .$
\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=\lim _{x \rightarrow 0}\left\{\frac{2 \sin ^2 \frac{m x}{2}}{2 \sin ^2 \frac{n x}{2}}\right\} \\
=\lim _{x \rightarrow 0}\left[\left\{\frac{\sin \frac{m x}{2}}{\frac{m x}{2}}\right\}^2 \frac{m^2 x^2}{4} \cdot \frac{1}{\left\{\frac{\sin \frac{n x}{2}}{\frac{n x}{2}}\right\}^2} \cdot \frac{4}{n^2 x^2}\right] \\
=\frac{m^2}{n^2} \times 1=\frac{m^2}{n^2} .
\end{gathered}$
Aliter : Apply L-Hospital's rule,
$\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=\lim _{x \rightarrow 0} \frac{m \sin m x}{n \sin n x}=\lim _{x \rightarrow 0} \frac{m^2 \cos m x}{n^2 \cos n x}=\frac{m^2}{n^2} .$
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