Question

$\underset{x\to -\infty }{lim}\frac{x^2\cdot \tan \frac{1}{x}}{\sqrt{8x^2+7x+1}}$

• A. $-\frac{1}{2\sqrt{2}}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{\sqrt{2}}$
• D. does not exist

Answer 1

The correct option is A $-\frac{1}{2\sqrt{2}}$

$\underset{x\to -\infty }{lim}\frac{x^2\times \tan \frac{1}{x}}{\sqrt{8x^2+7x+1}}$

Taking '$x^2$' common from denominator.

$\underset{x\to -\infty }{lim}\frac{x^2.\tan \frac{1}{x}}{|x|\sqrt{8+\frac{7}{x}+\frac{1}{x^2}}}$

$\underset{x\to -\infty }{lim}-\frac{\tan \frac{1}{x}}{\frac{1}{x}\sqrt{8+\frac{7}{x}+\frac{1}{x^2}}}$ ( $|x|=-x$ when $x<0$ )

$=\underset{x\to -\infty }{lim}-\frac{\sin \frac{1}{x}}{\frac{1}{x}\cos \left(\frac{1}{x}\right)\sqrt{8+\frac{7}{x}+\frac{1}{x^2}}}$

Let $\frac{1}{x}=t$

$\underset{t\to 0^{-}}{lim}-\frac{\sin t}{t\cos \left(t\right)\sqrt{8+7t+t^2}}$

$=-\frac{1}{2\sqrt{2}}$