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Q. The value of $\displaystyle\lim _{x \rightarrow-\infty} \frac{x^2 \tan \frac{1}{x}}{\sqrt{4 x^2-x+1}}$ is equal to

NTA AbhyasNTA Abhyas 2022

Solution:

Let $L=\displaystyle\lim _{x \rightarrow-\infty} \frac{x^2 \tan \frac{1}{x}}{\sqrt{4 x^2-x+1}}$
Taking out $x$ from square root in denominator, we get
$L =\displaystyle\lim _{x \rightarrow-\infty} \frac{x \frac{\tan \frac{1}{x}}{\frac{1}{x}}}{-x \sqrt{4-\frac{1}{x}+\frac{1}{x^2}}} $
$=\frac{1}{-\sqrt{4}} $
$=-\frac{1}{2}$