# $\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}$

Question.

$\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}$

solution:

Given $\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}$

Here cos mx can be written as

$\Rightarrow \cos m x=1-2 \sin ^{2} \frac{m x}{2}$

And similarly

$\Rightarrow \operatorname{cosn} x=1-2 \sin ^{2} \frac{n x}{2}$

Using these two identities in given equation we get

$\Rightarrow$$\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=\lim _{x \rightarrow 0} \frac{\left[1-\left(1-2 \sin ^{2} \frac{m x}{2}\right)\right]}{\left[1-\left(1-2 \sin ^{2} \frac{n x}{2}\right)\right]} On simplification we get \lim _{x \rightarrow 0} \frac{\left[1-\left(1-2 \sin ^{2} \frac{m x}{2}\right)\right]}{\left[1-\left(1-2 \sin ^{2} \frac{n x}{2}\right)\right]}=\lim _{x \rightarrow 0} \frac{2 \sin ^{2} \frac{m x}{2}}{2 \sin ^{2} \frac{2 \pi}{2}}=\lim _{x \rightarrow 0} \frac{\sin ^{2} \frac{m x}{2}}{\sin ^{2} \frac{1 x}{2}} Again by using trigonometric identity the above equation can be written as \Rightarrow$$\lim _{x \rightarrow 0} \frac{\sin ^{2} \frac{m x}{2}}{\sin ^{2} \frac{2 x}{2}}=\lim _{x \rightarrow 0} \frac{\left(\frac{m}{2}\right)^{2}\left[\frac{\sin ^{2} \frac{m x}{2}}{\left(\frac{m}{2}\right)^{2}}\right]}{\left(\frac{n}{2}\right)^{2}\left[\frac{\sin ^{2} \frac{n x}{2}}{\left(\frac{n}{2}\right)^{2}}\right]}$

Taking common and simplifying we get

$\Rightarrow$$\lim _{x \rightarrow 0} \frac{\left(\frac{m}{2}\right)^{2}\left[\frac{\sin ^{2} \frac{m x}{2}}{\left(\frac{m}{2}\right)^{2}}\right]}{\left(\frac{n}{2}\right)^{2}\left[\frac{\sin \frac{2 n x}{2}}{\left(\frac{n}{2}\right)^{2}}\right]}=\frac{\lim ^{2}}{n \rightarrow 0} \frac{\left[\frac{\sin ^{2} \frac{m x}{2}}{\left(\frac{m}{2}\right)^{2}}\right]}{\lim _{x \rightarrow 0}\left[\frac{\sin ^{2} \frac{n x}{2}}{\left(\frac{n}{2}\right)^{2}}\right]} Now as \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \Rightarrow$$\frac{\operatorname{m}^{2}}{\mathrm{n}^{2}} \frac{\lim _{\mathrm{x} \rightarrow 0}\left[\frac{\sin ^{2} \frac{\mathrm{mx}}{2}}{\left(\frac{\mathrm{m}}{2}\right)^{2}}\right]}{\lim _{\mathrm{x} \rightarrow 0}\left[\frac{\sin ^{2} \frac{\mathrm{n} \mathrm{x}}{2}}{\left(\frac{\mathrm{n}}{2}\right)^{2}}\right]}=\frac{\mathrm{m}^{2} 1}{\mathrm{n}^{2}} \frac{1}{1}=\frac{\mathrm{m}^{2}}{\mathrm{n}^{2}}$

By applying the limits we get

$\Rightarrow$$\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=\frac{m^{2}}{n^{2}}$