Functions Question 15
Question: $ \underset{x\to 0}{\mathop{\lim }}\frac{1-\cos mx}{1-\cos nx}= $
[Kerala (Engg.)2002]
Options:
A) $ m/n $
B) $ n/m $
C) $ \frac{m^{2}}{n^{2}} $
D) $ \frac{n^{2}}{m^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \underset{x\to 0}{\mathop{\lim }}\frac{1-\cos mx}{1-\cos nx}=\underset{x\to 0}{\mathop{\lim }}{ \frac{2{{\sin }^{2}}\tfrac{mx}{2}}{2{{\sin }^{2}}\tfrac{nx}{2}} } $ $ =\underset{x\to 0}{\mathop{\lim }}[ {{{ \frac{\sin \tfrac{mx}{2}}{\tfrac{mx}{2}} }}^{2}}\frac{m^{2}x^{2}}{4}.\frac{1}{{{{ \frac{\sin \tfrac{nx}{2}}{\tfrac{nx}{2}} }}^{2}}}.\frac{4}{n^{2}x^{2}} ] $ $ =\frac{m^{2}}{n^{2}}\times 1=\frac{m^{2}}{n^{2}} $ . Aliter : Apply L-Hospital?s rule, $ \underset{x\to 0}{\mathop{\lim }}\frac{1-\cos mx}{1-\cos nx}=\underset{x\to 0}{\mathop{\lim }}\frac{m\sin mx}{n\sin nx}=\underset{x\to 0}{\mathop{\lim }}\frac{m^{2}\cos mx}{n^{2}\cos nx}=\frac{m^{2}}{n^{2}}. $
BETA
