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

We can use the fact that for any polynomial function $P(x)$ of the same degree in the numerator and denominator, $\lim_{x\rightarrow \infty}\frac{P(x)}{Q(x)}=\frac{\text{leading coefficient of }P}{\text{leading coefficient of }Q}$. Therefore, $$\lim_{x\rightarrow\infty} \frac{1-6x}{1+8x}=\lim_{x\rightarrow\infty} \frac{-6x}{8x}=\boxed{\textbf{(B) } -\frac{3}{4}}.$$

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

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