Question

$\underset{x\to 0}{lim}\frac{\sin \left(x^2\right)}{\cos (x^2-x)}$ is equal to-

• A. $2$
• B. $-2$
• C. $1$
• D. $-1$

Answer 1

The correct option is C $1$

We have,

${lim}_{x\to 0}\frac{\sin \left(x^2\right)}{\cos (x^2-x)}$

Now,

${lim}_{x\to 0}\frac{x^2\times \frac{\sin \left(x^2\right)}{x^2}}{(x^2-x)\times \frac{\cos (x^2-x)}{(x^2-x)}}$

$\Rightarrow \frac{{lim}_{x\to 0}{x}^2\times {lim}_{x\to 0}\frac{\sin \left(x^2\right)}{x^2}}{{lim}_{x\to 0}(x^2-x)\times {lim}_{x\to 0}\frac{\cos (x^2-x)}{(x^2-x)}}$

$\Rightarrow {lim}_{x\to 0}\frac{x^2\times 1}{(x^2-x)\times 1}\therefore {lim}_{x\to 0}\frac{\sin x}{x}=1={lim}_{x\to 0}\frac{\cos x}{x}$

$\Rightarrow {lim}_{x\to 0}\frac{x^2}{x^2-x}\left(\frac{0}{0}form\right)$

Applying L’ Hospital rule and we get,

$\Rightarrow {lim}_{x\to 0}\frac{2x}{2x-1}$

$\Rightarrow {lim}_{x\to 0}\frac{2}{2-0}$

$\Rightarrow \frac{2}{2}$

$\Rightarrow 1$

Hence, this is the answer.