Question

$\underset{x\to 0}{lim}\frac{\sin 3x^2}{\ln \cos (2x^2-x)}$ is equal to :

• A. $0$
• B. $-6$
• C. $1$
• D. $\infty$

Answer 1

Answer: (A)

$0$

${lim}_{x\to 0}\frac{\sin {3x}^2}{\ln \left(\cos (2x^2-1)\right)}$
Applying L-Hospital Rule
${lim}_{x\to 0}$ $\frac{\cos \left({3x}^2\right).-6x}{\frac{(4x-1)}{\cos ({2x}^2-x)}}$
${lim}_{x\to 0}$ $\frac{\cos {3x}^2.6x}{(4x-1)}$
So, at $x$ tending to $0$,

We get $0$ as the answer.