# How do you determine the limit of sqrt(x-4)/(3x+5) as x approaches negative infinity?

Nov 11, 2017

$\frac{1}{3}$

#### Explanation:

Notice that you have a square root in your numerator. This can complicate the usual process, so you'll need to take a few extra steps to make sure this works.

The first step is the same as usual: we're going to divide everything by the highest power, which in this case is ${x}^{1}$:

${\lim}_{x \to - \infty} \frac{\frac{1}{x}}{\frac{1}{x}} \cdot \frac{\sqrt{x - 4}}{3 x + 5}$

For the numerator, we have to make one small simplification to get the right number into the square root:

$\implies {\lim}_{x \to - \infty} \frac{\frac{1}{\sqrt{{x}^{2}}}}{\frac{1}{x}} \cdot \frac{\sqrt{x - 4}}{3 x + 5}$

$\implies {\lim}_{x \to - \infty} \frac{\sqrt{\frac{x - 4}{x}}}{\frac{3 x + 5}{x}}$

Now, we divide through and simplify:

$\implies {\lim}_{x \to - \infty} \frac{\sqrt{\left(1 - \frac{4}{x}\right)}}{3 + \frac{5}{x}}$

$\implies \frac{\sqrt{\left(1 - \cancel{\frac{4}{x}}\right)}}{3 + \cancel{\frac{5}{x}}}$
$\implies \frac{1}{3}$