# What is  lim_(->-oo) f(x) = x^2/(x^2-7) -(2x^2)/(x^3-5)?

May 18, 2018

${\lim}_{x \to - \infty} f \left(x\right) = 1$

#### Explanation:

We'll take each of the two rational terms and divide through by their highest power of $x$. This gives:

f(x) = x^2 / (x^2 - 7) - (2x^2)/(x^3-5) = 1 / (1 - 7/x^2) - (2/x)/(1 - 5/x^3

Now recall the fact that ${\lim}_{x \to \pm \infty} \frac{a}{x} ^ n = 0$ if $a$ is a real number and $n \ge 1$. This tells us that $\frac{7}{x} ^ 2$, $\frac{2}{x}$ and $\frac{5}{x} ^ 3$ will all approach $0$ as $x \to - \infty$. Thus, we can directly replace these terms with zero which yields our limit.

${\lim}_{x \to - \infty} \frac{1}{1 - \frac{7}{x} ^ 2} - \frac{\frac{2}{x}}{1 - \frac{5}{x} ^ 3} = \frac{1}{1 - 0} - \frac{0}{1 - 0} = 1$.