Use l'H^opital's rule to find the limit.
-$\lim _{ x \rightarrow \infty } x \sin \frac { 16 } { x }$
A) 1
B) 0
C) 16
D) $\frac { 1 } { 16 }$

Question

Use l'H^opital's rule to find the limit.
-$\lim _{ x \rightarrow \infty }  x \sin \frac { 16 } { x }$
A) 1
B) 0
C) 16
D) $\frac { 1 } { 16 }$

Quizplus · Accepted Answer

We can apply l'Hopital's rule to evaluate this limit. We have:
\begin{align*}
\lim_{x	o\infty} x\sin\frac{16}{x} &= \lim_{x	o\infty} \frac{\sin\frac{16}{x}}{\frac{1}{x}}\
&= \lim_{x	o\infty} \frac{-\frac{16}{x^2}\cos\frac{16}{x}}{-\frac{1}{x^2}}\
&= \lim_{x	o\infty} 16\cos\frac{16}{x} \
&= 16\cos 0\
&= \boxed{16}
\end{align*}

Use L'H^opital's Rule to Find the Limit lim⁡x→∞xsin⁡16x\lim _{ x \rightarrow \infty } x \sin \frac { 16 } { x }limx→∞​xsinx16​

Use L'H^opital's Rule to Find the Limit $\lim _{ x \rightarrow \infty } x \sin \frac { 16 } { x }$