Question

$\underset{x\to -\infty }{\lim }\frac{x^5\tan \left(\frac{1}{\pi x^2}\right)+3|x{|}^2+7}{|x{|}^3+7|x|+8}$ is equal to

• A. $-\frac{1}{\pi }$
• B. 0
• C. $\infty$
• D. does not exist

Answer 1

Answer: (A)

We have, $\underset{x\to -\infty }{\lim }\frac{x^5\tan \left(\frac{1}{\pi x^2}\right)+3|x{|}^2+7}{|x{|}^3+7|x|+8}$
$=\underset{x\to -\infty }{\lim }\frac{x^5\tan \left(\frac{1}{\pi x^2}\right)+3x^2+7}{-x^3-7x+8}$ $[\because x<0\Rightarrow |x|=-x]$
$=\underset{x\to -\infty }{\lim }\frac{x^2\tan \left(\frac{1}{\pi x^2}\right)+\frac{3}{x}+\frac{7}{x^3}}{-1-\frac{7}{x^2}+\frac{8}{x^3}}$
On dividing the numerator and denominator by $x^3$
$=\underset{x\to -\infty }{\lim }\frac{\frac{1}{\pi }\cdot \frac{\tan \left\{\frac{1}{\left(\pi x^2\right)}\right\}}{\frac{1}{\left(\pi x^2\right)}}+\frac{3}{x}+\frac{7}{x^3}}{-1-\frac{7}{x^2}+\frac{8}{x^3}}$
$=\frac{\frac{1}{\pi }\cdot 1+0+0}{-1-0+0}=-\frac{1}{\pi }$