Differentiation Question 36
Question: $ \underset{x\to 0}{\mathop{\lim }}\frac{\sin (x^{2})}{in(cos(2x^{2}-x))} $ is equal to
Options:
A) 2
B) -2
C) 1
D) -1
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \underset{x\to 0}{\mathop{\lim }}\frac{\sin (x^{2})}{In(cos(2x^{2}-x))} $
$ =\underset{x\to 0}{\mathop{\lim }}\frac{\sin (x^{2})}{\log ( 1-2{{\sin }^{2}}( \frac{2x^{2}-x}{2} ) )} $
$ =\underset{x\to 0}{\mathop{\lim }}\frac{\sin (x^{2})x^{2}}{\frac{x^{2}\log ( 1-2{{\sin }^{2}}( \frac{2x^{2}-x}{2} ) )}{-2{{\sin }^{2}}( \frac{2x^{2}-x}{2} )}[ -2{{\sin }^{2}}( \frac{2x^{2}-x}{2} ) ]}= $
$ \underset{x\to 0}{\mathop{\lim }}\frac{x^{2}}{\frac{2{{\sin }^{2}}( \frac{2x^{2}-x}{2} )}{{{( \frac{2x^{2}-x}{2} )}^{2}}}{{( \frac{2x^{2}-x}{2} )}^{2}}} $ = $ \underset{x\to 0}{\mathop{\lim }}-\frac{2x^{2}}{{{(2x^{2}-x)}^{2}}}=\underset{x\to 0}{\mathop{\lim }}-\frac{2}{{{(2x-1)}^{2}}}=-2 $
BETA
