Find $\lim _ { x \rightarrow \infty } \frac { 1 - 3 x } { 1 + 7 x }$ (if it exists).
A) 0
B) $- \frac { 3 } { 7 }$
C) $\frac { 3 } { 7 }$
D) $- \infty$
E) limit does not exist

Question

Quizplus · Accepted Answer

To apply L'Hôpital's rule, we take the derivative of the numerator and the denominator with respect to x:
$$\lim_{x \rightarrow \infty} \frac{1-3x}{1+7x} = \lim_{x \rightarrow \infty}\frac{-3}{7}\cdot \frac {x}{x} = -\frac{3}{7}$$
Thus, the answer is (B) $- \frac { 3 } { 7 }$.

Find lim⁡x→∞1−3x1+7x\lim _ { x \rightarrow \infty } \frac { 1 - 3 x } { 1 + 7 x }limx→∞​1+7x1−3x​

Find $\lim _ { x \rightarrow \infty } \frac { 1 - 3 x } { 1 + 7 x }$