Find $\lim _ { x \rightarrow \infty } \frac { 1 - 7 x } { 1 + 9 x }$ (if it exists).
A) 0
B) $- \frac { 7 } { 9 }$
C) $\frac { 7 } { 9 }$
D) $- \infty$
E) limit does not exist

Question

Quizplus · Accepted Answer

We can apply the limit rules for rational functions (divide numerator and denominator by the highest power of x and then take the limit as x goes to infinity):
$$\lim _ { x \rightarrow \infty } \frac { 1 - 7 x } { 1 + 9 x } = \lim _ { x \rightarrow \infty } \frac { \frac { 1 } { x } - 7 } { \frac { 1 } { x } + 9 } = \frac { \lim _ { x \rightarrow \infty } \frac { 1 } { x } - 7 } { \lim _ { x \rightarrow \infty } \frac { 1 } { x } + 9 } = \frac{0-7}{0+9} = -\frac{7}{9}$$
Therefore, the answer is (B): $-\frac{7}{9}$.

Find lim⁡x→∞1−7x1+9x\lim _ { x \rightarrow \infty } \frac { 1 - 7 x } { 1 + 9 x }limx→∞​1+9x1−7x​

Find $\lim _ { x \rightarrow \infty } \frac { 1 - 7 x } { 1 + 9 x }$