Question

$\underset{x\to -\infty }{\mathrm{Lim}}\frac{x^2\tan \left(\frac{1}{x}\right)-x}{1-|x|}$ equals

• A. 0
• B. -1
• C. 1
• D. does not exist

Answer 1

Answer: (A)

$\underset{x\to -\infty }{\mathrm{Lim}}\frac{x^2\tan \left(\frac{1}{x}\right)-x}{1-|x|}$
Put $x=\frac{1}{y}$,
so $\underset{y\to 0}{\mathrm{Lim}}\frac{\frac{1}{y^2}\cdot \tan y-\frac{1}{y}}{1+\frac{1}{y}}=\underset{y\to 0}{\mathrm{Lim}}\frac{\left(\frac{\tan y}{y}-1\right)}{y+1}$
$=\frac{1-1}{0+1}=0$.