Find the limit. $\lim _ { x \rightarrow \infty } \frac { - 8 x + 2 } { - 5 x ^ { 2 } + 4 }$
A) $\frac { 1 } { 2 }$
B) 1
C) 0
D) $\infty$
E) $\frac { 8 } { 5 }$

Question

Quizplus · Accepted Answer

As $x$ approaches infinity, the denominator $-5x^2+4$ goes to negative infinity, while the numerator $-8x+2$ goes to negative infinity faster. Thus, the entire fractions goes to $\frac{-\infty}{-\infty}$, which is an indeterminate form. Therefore, we can use L'Hopital's rule by taking the derivative of the numerator and denominator with respect to $x$. 
\begin{align*}
\lim_{x	o\infty}\frac{-8x+2}{-5x^2+4}&=\lim_{x	o\infty}\frac{\frac{d}{dx}(-8x+2)}{\frac{d}{dx}(-5x^2+4)}\
&=\lim_{x	o \infty}\frac{-8}{-10x}\
&=\lim_{x	o\infty}\frac{4}{5x}= \boxed{	extbf{(C) }0}.
\end{align*}

Find the Limit lim⁡x→∞−8x+2−5x2+4\lim _ { x \rightarrow \infty } \frac { - 8 x + 2 } { - 5 x ^ { 2 } + 4 }limx→∞​−5x2+4−8x+2​

Find the Limit $\lim _ { x \rightarrow \infty } \frac { - 8 x + 2 } { - 5 x ^ { 2 } + 4 }$