Question

find the following limit. notes: enter \dne\ if limit does not exist. $lim_{x
ightarrowinfty}\frac{-1 + 6e^{x}}{-2 + 3e^{x}}=square$

Explanation:

Step1: Divide numerator and denominator by $e^{x}$

As $x\to\infty$, we know that $\lim_{x\to\infty}\frac{1}{e^{x}} = 0$. Divide the numerator and denominator of $\frac{-1 + 6e^{x}}{-2+3e^{x}}$ by $e^{x}$: $\lim_{x\to\infty}\frac{\frac{-1}{e^{x}}+6}{\frac{-2}{e^{x}} + 3}$.

Step2: Evaluate the limit

Since $\lim_{x\to\infty}\frac{-1}{e^{x}}=0$ and $\lim_{x\to\infty}\frac{-2}{e^{x}} = 0$, we have $\frac{\lim_{x\to\infty}\frac{-1}{e^{x}}+6}{\lim_{x\to\infty}\frac{-2}{e^{x}}+3}=\frac{0 + 6}{0+3}$.

Step3: Simplify the result

$\frac{6}{3}=2$.

Answer:

$2$