$\text { Evaluate the limit } \lim _ { x \rightarrow \infty } \frac { 56 - 7 x + 8 x ^ { 2 } } { 5 x ^ { 2 } - 7 } \text { first by using techniques from Chapter } 3 \text { then }$ by using L'Hopital's Rule.
A) $\frac { 56 } { 5 }$
B) $\frac { 8 } { 5 }$
C) $- \frac { 7 } { 5 }$
D) $\infty$
E) does not exist

Question

$\text { Evaluate the limit } \lim _ { x \rightarrow \infty } \frac { 56 - 7 x + 8 x ^ { 2 } } { 5 x ^ { 2 } - 7 } \text { first by using techniques from Chapter } 3 \text { then }$ by using L'Hopital's Rule. 
A) $\frac { 56 } { 5 }$
B) $\frac { 8 } { 5 }$
C) $- \frac { 7 } { 5 }$
D) $\infty$
E) does not exist

Quizplus · Accepted Answer

To evaluate the limit as $x$ approaches infinity, divide every term by the highest power of $x$ in the denominator, which is $x^2$. This simplifies to $\lim _ { x \rightarrow \infty } \frac { 8 + \frac{56}{x^2} - \frac{7}{x} } { 5 - \frac{7}{x^2} }$, which approaches $\frac{8}{5}$ as $x$ goes to infinity. Using L'Hopital's Rule, which applies because the original form is an indeterminate form ($\frac{\infty}{\infty}$), differentiating the numerator and denominator separately also leads to the limit $\frac{8}{5}$ as $x$ approaches infinity.

By Using L'Hopital's Rule 565\frac { 56 } { 5 }556​

By Using L'Hopital's Rule $\frac { 56 } { 5 }$