Question

${lim}_{x\to 0}\frac{\sin \left(6x^2\right)}{ln\cos (2x^2-x)}$ is equal to

• A. $12$
• B. $-12$
• C. $6$
• D. $-6$

Answer 1

The correct option is B $-12$

We have,

${lim}_{x\to 0}\left(\frac{\sin \left(6x^2\right)}{\ln \left(\cos (2x^2-x)\right)}\right)$

This is the $\frac{0}{0}$ form.

So, apply L-Hospital rule

${lim}_{x\to 0}\left(\frac{\cos \left(6x^2\right)\times \left(12x\right)}{\frac{1}{\cos (2x^2-x)}\times (-\sin (2x^2-x))\times (4x-1)}\right)$

${lim}_{x\to 0}\left(\frac{12x\cos \left(6x^2\right)}{-\tan (2x^2-x)\times (4x-1)}\right)$

This is the $\frac{0}{0}$ form.

So, apply L-Hospital rule

${lim}_{x\to 0}\left(\frac{12x\cos \left(6x^2\right)}{-(4x-1)\tan (2x^2-x)}\right)$

${lim}_{x\to 0}\left(\frac{12\cos \left(6x^2\right)+12x(-\sin \left(6x^2\right))\times 12x}{-(4x-1){\sec }^2(2x^2-x)\times (4x-1)-\tan (2x^2-x)(4-0)}\right)$

${lim}_{x\to 0}\left(\frac{12\cos \left(6x^2\right)-144x^2\sin \left(6x^2\right)}{-{(4x-1)}^2{\sec }^2(2x^2-x)-4\tan (2x^2-x)}\right)$

$=\frac{12\cos 0-0}{-{(0-1)}^2{\sec }^{2}0-4\tan 0}$

$=\frac{12}{-1-0}$

$=-12$

Hence, this is the answer.