Question

determine whether the following limit is equal to $infty$, $-infty$ or some specific value. $lim_{x
ightarrow-infty}\frac{1 + e^{x}}{x^{2}+6}$

Explanation:

Step1: Analyze the limit of numerator

As $x\to -\infty$, we know that $\lim_{x\to -\infty}e^{x}=0$. So $\lim_{x\to -\infty}(1 + e^{x})=1+0 = 1$.

Step2: Analyze the limit of denominator

As $x\to -\infty$, $\lim_{x\to -\infty}(x^{2}+6)=\lim_{x\to -\infty}x^{2}+6$. Since $\lim_{x\to -\infty}x^{2}=\infty$, then $\lim_{x\to -\infty}(x^{2}+6)=\infty$.

Step3: Use limit - quotient rule

By the quotient - rule for limits $\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$ (where $\lim_{x\to a}g(x)
eq0$), we have $\lim_{x\to -\infty}\frac{1 + e^{x}}{x^{2}+6}=\frac{\lim_{x\to -\infty}(1 + e^{x})}{\lim_{x\to -\infty}(x^{2}+6)}=\frac{1}{\infty}=0$.

Answer:

$0$