# What is the limit of (x^2 + x- 30)/|x- 5| as x approaches infinity?

Mar 8, 2015

${\lim}_{x \rightarrow \infty} \frac{{x}^{2} + x - 30}{\left\mid x - 5 \right\mid} = \infty$

For positive values of $x$, $\left\mid x - 5 \right\mid = x - 5$. So,

${\lim}_{x \rightarrow \infty} \frac{{x}^{2} + x - 30}{\left\mid x - 5 \right\mid} = {\lim}_{x \rightarrow \infty} \frac{{x}^{2} + x - 30}{x - 5}$.

This limit evaluates to the indeterminate form $\frac{\infty}{\infty}$.

(If you have access to l'Hopital's rule, you could use it, but it is not necessary.)

Use algebra to make the denominator not go to $\infty$.

For all positive $x$ (which is all we're interested in as $x \rightarrow \infty$),

$\frac{{x}^{2} + x - 30}{\left\mid x - 5 \right\mid} = \frac{{x}^{2} + x - 30}{x - 5} = \frac{x \left(x + 1 - \frac{30}{x}\right)}{x \left(1 - \frac{5}{x}\right)} = \frac{x + 1 - \frac{30}{x}}{1 - \frac{5}{x}}$

As $x$ increases without bound, the numerator of this expression also increases without bound and the denominator approaches $1$.

Therefore:${\lim}_{x \rightarrow \infty} \frac{{x}^{2} + x - 30}{\left\mid x - 5 \right\mid} = {\lim}_{x \rightarrow \infty} \frac{x + 1 - \frac{30}{x}}{1 - \frac{5}{x}} = \infty$