Question

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

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

Answer 1

Answer: (A)

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