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

To find the limit as $ x $ approaches infinity, we must look at the behavior of the numerator and denominator as $ x $ gets very large. Since the coefficient of $ x $ in the numerator is negative and the coefficient of $ x $ in the denominator is positive, the numerator approaches negative infinity and the denominator approaches positive infinity as $ x $ goes to infinity. Therefore, we can apply L'Hôpital's rule and take the derivative of both the numerator and denominator with respect to $ x $:
$$\lim_{x\rightarrow \infty} \frac{1-7x}{1+9x} = \lim_{x\rightarrow \infty} \frac{-7}{9} = -\frac{7}{9}$$
So the answer is B.

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

Find $\lim _{x \rightarrow \infty} \frac{1-7 x}{1+9 x}$ (If It Exists)