Find $\lim _ { x \rightarrow \infty } \frac { 1 - 6 x } { 1 + 8 x }$ (if it exists).
A) 0
B) $- \frac { 3 } { 4 }$
C) $\frac { 3 } { 4 }$
D) $- \infty$
E) limit does not exist

Question

Quizplus · Accepted Answer

If we divide numerator and denominator by $x$, we get
$$\lim_{x ightarrow \infty} \frac{\frac{1}{x} - 6}{\frac{1}{x} + 8}$$
Since both the numerator and denominator go to $0$ as $x ightarrow \infty$, we can apply L'Hopital's Rule
$$\lim_{x ightarrow \infty} \frac{\frac{1}{x}-6}{\frac{1}{x}+8} = \lim_{x ightarrow \infty} \frac{-\frac{1}{x^2}}{-\frac{1}{x^2}} \cdot \frac{1}{-\frac{1}{x^2}} = \lim_{x ightarrow \infty} \frac{1}{\frac{1}{x^2}} \cdot x^2 \cdot \frac{-6}{8} = \frac{-3}{4}$$
Thus, the answer is $\boxed{	extbf{(B)}}$.

Find lim⁡x→∞1−6x1+8x\lim _ { x \rightarrow \infty } \frac { 1 - 6 x } { 1 + 8 x }limx→∞​1+8x1−6x​

Find $\lim _ { x \rightarrow \infty } \frac { 1 - 6 x } { 1 + 8 x }$