$\text { Evaluate the } \lim _ { x \rightarrow \infty } \frac { 2 x - 6 } { 4 x ^ { 2 } + 8 x + 5 } \text { using L'Hôpital's Rule if necessary. }$
A) $\frac { 1 } { 6 }$
B) $\frac { 1 } { 4 }$
C) $\frac { 1 } { 2 }$
D) $\infty$
E) 0

Question

$\text { Evaluate the } \lim _ { x \rightarrow \infty } \frac { 2 x - 6 } { 4 x ^ { 2 } + 8 x + 5 } \text { using L'Hôpital's Rule if necessary. }$ 
A) $\frac { 1 } { 6 }$
B) $\frac { 1 } { 4 }$
C) $\frac { 1 } { 2 }$
D) $\infty$
E) 0

Quizplus · Accepted Answer

By looking at the degree of the numerator and denominator, we see that the denominator grows faster and hence the limit is 0. L'Hopital's Rule is not necessary in this case. However, if we were to use L'Hopital's Rule, we would get the limit as $\lim_{x	o\infty}\frac{2}{8x+8}=0$.