Question

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

• A. 2
• B. -2
• C. 1
• D. -1

Answer 1

Answer: (B)

-2

$\underset{x\to 0}{lim}\frac{\sin \left(x^2\right)}{\ln \left(\cos (2x^2-x)\right)}$$=\underset{x\to 0}{lim}\frac{\sin \left(x^2\right)}{\log (1-2{\sin }^2\left(\frac{2x^2-x}{2}\right))}$
$=\underset{x\to 0}{lim}\frac{\sin \left(x^2\right)x^2}{\frac{x^2\log (1-2{\sin }^2\left(\frac{2x^2-x}{2}\right))}{-2{\sin }^2\left(\frac{2x^2-x}{2}\right)}[-2{\sin }^2\left(\frac{2x^2-x}{2}\right)]}$
$=\underset{x\to 0}{lim}-\frac{x^2}{\frac{2{\sin }^2\left(\frac{2x^2-x}{2}\right)}{{\left(\frac{2x^2-x}{2}\right)}^2}{\left(\frac{2x^2-x}{2}\right)}^2}$
$=\underset{x\to 0}{lim}-\frac{2x^2}{(2x^2-x{)}^2}=\underset{x\to 0}{lim}-\frac{2}{(2x-1{)}^2}=-2$