Question

$\underset{x\to -\infty }{lim}\frac{x^4\sin \left(\frac{1}{x}\right)+x^2}{1+|x{|}^3}=$

• A. $0$
• B. $\infty$
• C. $1$
• D. $-1$

Answer 1

The correct option is D $-1$

${lim}_{x\to -\infty }\frac{x^{4}sin\left(\frac{1}{x}\right)+x^2}{1+|x{|}^3}=?$

Since $x\to -\infty$, $x$ is negative. Therefore $|x|=-x$

${lim}_{x\to -\infty }\frac{x^{4}sin\left(\frac{1}{x}\right)+x^2}{1+|x{|}^3}={lim}_{x\to -\infty }\frac{x^{4}sin\left(\frac{1}{x}\right)+x^2}{1-x^3}$

Divide the numerator and denominator by $x^3$

$=li{m}_{x\to -\infty }\frac{xsin\left(\frac{1}{x}\right)+\frac{1}{x}}{\frac{1}{x^3}-1}$

$=\frac{{lim}_{x\to -\infty }(xsin\left(\frac{1}{x}\right)+\frac{1}{x})}{{lim}_{x\to -\infty }(\frac{1}{x^3}-1)}$

$=\frac{{lim}_{x\to -\infty }\left(xsin\left(\frac{1}{x}\right)\right)+{lim}_{x\to -\infty }\left(\frac{1}{x}\right)}{{lim}_{x\to -\infty }\left(\frac{1}{x^3}\right)-{lim}_{x\to -\infty }\left(1\right)}$

$=\frac{{lim}_{x\to -\infty }\left(\frac{sin\left(\frac{1}{x}\right)}{\frac{1}{x}}\right)+{lim}_{x\to -\infty }\left(\frac{1}{x}\right)}{{lim}_{x\to -\infty }\left(\frac{1}{x^3}\right)-{lim}_{x\to -\infty }\left(1\right)}$

Apply L'hopital's rule to the first term in the numerator.

$=\frac{{lim}_{x\to -\infty }\left(\frac{\frac{-cos\left(\frac{1}{x}\right)}{x^2}}{\frac{-1}{x^2}}\right)+{lim}_{x\to -\infty }\left(\frac{1}{x}\right)}{{lim}_{x\to -\infty }\left(\frac{1}{x^3}\right)-{lim}_{x\to -\infty }\left(1\right)}$

$=\frac{1+0}{0-1}$

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

$=-1$

$\therefore {lim}_{x\to -\infty }\frac{x^{4}sin\left(\frac{1}{x}\right)+x^2}{1+|x{|}^3}=-1$