Question #8da62

Feb 3, 2018

First, divide both the numerator and denominator by ${x}^{2}$.

${\lim}_{x \to - \infty} \frac{- 2 {x}^{5} + {x}^{4} - 3}{3 {x}^{2} - 7}$

${\lim}_{x \to - \infty} \frac{- 2 {x}^{3} + {x}^{2} - \frac{3}{x} ^ 2}{3 - \frac{7}{x} ^ 2}$

The $\frac{3}{x} ^ 2$ and $\frac{7}{x} ^ 2$ terms will go to $0$ as $x \to \infty$, so we can eliminate them from the expression.

${\lim}_{x \to - \infty} \frac{- 2 {x}^{3} + {x}^{2}}{3}$

Now, we can prove that this limit equals $\infty$.

Since ${x}^{3}$ is an odd power, it will tend towards $- \infty$ as $x \to - \infty$.
Therefore, multiplying ${x}^{3}$ by a negative number will make it tend towards $+ \infty$.

Therefore, $- 2 {x}^{3}$ will tend towards $+ \infty$ as $x \to - \infty$.

Since ${x}^{2}$ is an even power, it will tend towards $+ \infty$ as $x \to - \infty$

Since BOTH of the top terms will tend towards $+ \infty$, and the bottom term remains constant, the entire fraction will tend towards $+ \infty$. In other words:

${\lim}_{x \to - \infty} \frac{- 2 {x}^{3} + {x}^{2}}{3} = + \infty$

$\therefore {\lim}_{x \to - \infty} \frac{- 2 {x}^{5} + {x}^{4} - 3}{3 {x}^{2} - 7} = + \infty$