Question

determine whether the following limit is equal to $infty$, $-infty$ or some specific value. $lim_{x
ightarrow-infty}\frac{1 - 6x^{2}}{4x^{2}-e^{-x}}$

Explanation:

Step1: Divide numerator and denominator by $x^{2}$

As $x\to -\infty$, we divide both the numerator and denominator of $\frac{1 - 6x^{2}}{4x^{2}-e^{-x}}$ by $x^{2}$. We get $\lim_{x\to -\infty}\frac{\frac{1}{x^{2}}-6}{4-\frac{e^{-x}}{x^{2}}}$.

Step2: Analyze the limit of each term

We know that $\lim_{x\to -\infty}\frac{1}{x^{2}} = 0$. For $\lim_{x\to -\infty}\frac{e^{-x}}{x^{2}}$, as $x\to -\infty$, let $t=-x$, then $t\to\infty$ and we have $\lim_{t\to\infty}\frac{e^{t}}{t^{2}}$. Using L - H rule twice: $\lim_{t\to\infty}\frac{e^{t}}{t^{2}}=\lim_{t\to\infty}\frac{e^{t}}{2t}=\lim_{t\to\infty}\frac{e^{t}}{2}=\infty$. But since we are considering $\frac{e^{-x}}{x^{2}}$ as $x\to -\infty$, $\lim_{x\to -\infty}\frac{e^{-x}}{x^{2}}=\infty$. However, when we consider the whole fraction $\frac{\frac{1}{x^{2}}-6}{4 - \frac{e^{-x}}{x^{2}}}$, we can also note that as $x\to -\infty$, the dominant terms in the numerator and denominator are $- 6x^{2}$ and $4x^{2}$ respectively.
Another way is to directly consider the dominant terms. As $x\to -\infty$, the term $e^{-x}\to\infty$ but the term $x^{2}$ in the denominator and numerator grows much slower in comparison to the $x^{2}$ - terms. The dominant terms of the numerator and denominator are $-6x^{2}$ and $4x^{2}$. So $\lim_{x\to -\infty}\frac{1 - 6x^{2}}{4x^{2}-e^{-x}}=\lim_{x\to -\infty}\frac{-6x^{2}}{4x^{2}}$.

Step3: Calculate the limit

$\lim_{x\to -\infty}\frac{-6x^{2}}{4x^{2}}=-\frac{6}{4}=-\frac{3}{2}$

Answer:

$-\frac{3}{2}$