Question

determine whether the following limit is equal to $infty$, $-infty$ or some specific value. $lim_{x
ightarrow-infty}\frac{2x^{2}+4}{6x^{2}-e^{x}}$

Explanation:

Step1: Divide by highest - power of x in denominator

Divide both the numerator and denominator by $x^{2}$. We get $\lim_{x
ightarrow-\infty}\frac{2 + \frac{4}{x^{2}}}{6-\frac{e^{x}}{x^{2}}}$.

Step2: Analyze the limit of each term

As $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}\frac{4}{x^{2}} = 0$ since the denominator $x^{2}
ightarrow+\infty$ as $x
ightarrow\pm\infty$. Also, for $\lim_{x
ightarrow-\infty}\frac{e^{x}}{x^{2}}$, we know that $y = e^{x}$ approaches 0 as $x
ightarrow-\infty$ and $x^{2}
ightarrow+\infty$ as $x
ightarrow-\infty$, so $\lim_{x
ightarrow-\infty}\frac{e^{x}}{x^{2}}=0$.

Step3: Calculate the overall limit

Using the limit rules for quotients $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x
ightarrow a}f(x)}{\lim_{x
ightarrow a}g(x)}$ (where $\lim_{x
ightarrow a}g(x)
eq0$), we have $\frac{\lim_{x
ightarrow-\infty}(2 + \frac{4}{x^{2}})}{\lim_{x
ightarrow-\infty}(6-\frac{e^{x}}{x^{2}})}=\frac{2 + 0}{6-0}$.

Answer:

$\frac{1}{3}$