Question

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

(a) $\left(\frac{m}{n}\right)^2$

(b) $\left(\frac{n}{m}\right)^2$

(c) $\frac{n}{m}$

(d) $\frac{m}{n}

Answer 1

$\lim _{x \rightarrow 0} \frac{1-\cos n x}{1-\cos m x}\left(\frac{0}{0}\right.$ form$)=\lim _{x \rightarrow 0} \frac{n \sin n x}{m \sin m x}\left(\frac{0}{0}\right.$ form$)$

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

So, The correct option of this question will be (C).