Question

$\underset{x\to 0}{lim}\frac{\sin x^{2012}-x^{2012}\cos x^{2012}+(x^3{)}^{2012}}{x^{2012}(e^{5.x^{2012}}-1-5.x^{2012})}$ is equal to-

• A. $-\frac{1}{3}$
• B. $\frac{1}{3}$
• C. $\frac{4}{3}$
• D. $\frac{8}{75}$

Answer 1

The correct option is D $\frac{8}{75}$

Let $x^{2012}=t$ then $t\to 0asx\to 0$
$\therefore$ Given limit = $\underset{x\to 0}{lim}\frac{\sin t-t\cos t+t^3}{t(e^{5t}-1-5t)}$

= $\underset{t\to 0}{lim}\frac{\sin t-t\cos t+t^3}{t.25t^2\left(\frac{e^{5t}-1-5t}{25t^2}\right)}$

(use standard limit $\underset{x\to 0}{lim}\frac{e^x-1-x}{x^2}=\frac{1}{2}$)

= $\frac{2}{25}\underset{t\to 0}{lim}\frac{\sin t-t\cos t+t^3}{t^3}$

=$\frac{2}{25}\underset{t\to 0}{lim}\frac{(t-\frac{t^3}{3!}+\frac{t^5}{5!}...)-t(1-\frac{t^2}{2!}+\frac{t^4}{4!}...)+t^3}{t^3}$

= $\frac{2}{25}.(\frac{1}{2}+1-\frac{1}{6})=\frac{8}{75}$