Question

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

Explanation:

Step1: Analyze the behavior of each term as $x\to -\infty$

As $x\to -\infty$, $x^{2}\to+\infty$, $e^{x}\to0$, and $4x - 1\to-\infty$.

Step2: Consider the dominant - term

The dominant term in the numerator as $x\to -\infty$ is $x^{2}$ (since $e^{x}\to0$). We can use the fact that for large - magnitude negative $x$, the function behaves like $\frac{x^{2}}{4x}=\frac{1}{4}x$.

Step3: Determine the limit

Since $\lim_{x\to -\infty}\frac{x^{2}}{4x - 1}=\lim_{x\to -\infty}\frac{x^{2}/x}{(4x - 1)/x}=\lim_{x\to -\infty}\frac{x}{4-\frac{1}{x}}$. As $x\to -\infty$, $\frac{1}{x}\to0$, and $\lim_{x\to -\infty}\frac{x}{4 - \frac{1}{x}}=-\infty$.

Answer:

$-\infty$