Question

$\underset{x\to 0}{lim}\frac{1-\cos mx}{1-\cos nx}$=

• A. $m^2$
• B. $n^2$
• C. $m^2-n^2$
• D. $\frac{m^2}{n^2}$

Answer 1

Answer: (D)

$\frac{m^2}{n^2}$

Using the identity,
$\cos 2x=1-{\sin }^{2}x$
The given expression can be written in the form of,
$\underset{x\to 0}{lim}\frac{2{\sin }^2\left(\frac{mx}{2}\right)}{2{\sin }^2\left(\frac{nx}{2}\right)}$
$=\underset{x\to 0}{lim}\frac{2{sin}^2\left(\frac{mx}{2}\right)}{{\left(\frac{mx}{2}\right)}^2}\times \frac{{\left(\frac{nx}{2}\right)}^2}{2{sin}^2\left(\frac{nx}{2}\right)}\times \frac{m^2}{n^2}$
Using, $limx\to 0\frac{sinx}{x}=1$
Answer $=\frac{m^2}{n^2}$