Question

question 3 - 1 point evaluate the following limit using lhospitals rule. enter an exact answer. provide your answer below: $lim_{x \to infty}\frac{11x + 2}{8x - 7}$ $lim_{x \to infty}\frac{11x + 2}{8x - 7}=square$

Explanation:

Step1: Check form of limit

As $x\to\infty$, the limit $\lim_{x\to\infty}\frac{11x + 2}{8x-7}$ is in the $\frac{\infty}{\infty}$ form, so L'Hospital's rule can be applied.

Step2: Differentiate numerator and denominator

The derivative of the numerator $u = 11x+2$ is $u'=11$, and the derivative of the denominator $v = 8x - 7$ is $v'=8$.

Step3: Find new limit

By L'Hospital's rule, $\lim_{x\to\infty}\frac{11x + 2}{8x-7}=\lim_{x\to\infty}\frac{11}{8}$.

Answer:

$\frac{11}{8}$