Question

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

• A. $-1/\pi$
• B. 0
• C. $\infty$
• D. Does not exist

Answer 1

The correct option is A

$-1/\pi$

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

Since $[x\to -\infty \Rightarrow \frac{1}{\pi x^2}\to 0and\underset{\theta \to 0}{lim}\frac{tan\theta }{\theta }=1]=\frac{\frac{1}{\pi }.1+0+0}{-1-0+0}=-\frac{1}{\pi }$