Question

determine whether the following limit is equal to $infty$, $-infty$ or some specific value. $lim_{x
ightarrow-infty}\frac{5x - e^{-x}}{4x^{3}+2x}$

Explanation:

Step1: Divide by highest - power of x in denominator

Divide both the numerator and denominator by $x^{3}$.
The expression becomes $\lim_{x
ightarrow-\infty}\frac{\frac{5x}{x^{3}}-\frac{e^{-x}}{x^{3}}}{\frac{4x^{3}}{x^{3}}+\frac{2x}{x^{3}}}=\lim_{x
ightarrow-\infty}\frac{\frac{5}{x^{2}}-\frac{e^{-x}}{x^{3}}}{4 + \frac{2}{x^{2}}}$.

Step2: Analyze the limit of each term

• For $\lim_{x

ightarrow-\infty}\frac{5}{x^{2}}$, as $x
ightarrow-\infty$, $x^{2}
ightarrow\infty$, so $\lim_{x
ightarrow-\infty}\frac{5}{x^{2}} = 0$.

• For $\lim_{x

ightarrow-\infty}\frac{2}{x^{2}}$, as $x
ightarrow-\infty$, $x^{2}
ightarrow\infty$, so $\lim_{x
ightarrow-\infty}\frac{2}{x^{2}} = 0$.

• For $\lim_{x

ightarrow-\infty}\frac{e^{-x}}{x^{3}}$, use L'Hopital's rule multiple - times. Let $y=\frac{e^{-x}}{x^{3}}$, then $\lim_{x
ightarrow-\infty}y=\lim_{x
ightarrow-\infty}\frac{e^{-x}}{x^{3}}=\lim_{x
ightarrow-\infty}\frac{-e^{-x}}{3x^{2}}=\lim_{x
ightarrow-\infty}\frac{e^{-x}}{6x}=\lim_{x
ightarrow-\infty}\frac{-e^{-x}}{6}=-\infty$.

Step3: Calculate the overall limit

$\lim_{x
ightarrow-\infty}\frac{\frac{5}{x^{2}}-\frac{e^{-x}}{x^{3}}}{4+\frac{2}{x^{2}}}=\frac{0-(-\infty)}{4 + 0}=\infty$.

Answer:

$\infty$