# How do you find the limit of 1/(x^3 +4) as x approaches oo?

Feb 19, 2017

${\lim}_{x \rightarrow \infty} \frac{1}{{x}^{3} + 4} = 0$

#### Explanation:

We can multiply numerator and denominator by $\frac{1}{x} ^ 3$ (the reciprocal of the largest power in the denominator) as follows:

${\lim}_{x \rightarrow \infty} \frac{1}{{x}^{3} + 4} = {\lim}_{x \rightarrow \infty} \frac{\left(\frac{1}{x} ^ 3\right) \left(1\right)}{\left(\frac{1}{x} ^ 3\right) \left({x}^{3} + 4\right)}$
$\text{ } = {\lim}_{x \rightarrow \infty} \frac{\frac{1}{x} ^ 3}{1 + \frac{4}{x} ^ 3}$

And we note that as $x \rightarrow \infty$ then $\frac{1}{x} , \frac{1}{x} ^ 2 , \frac{1}{x} ^ 3 \rightarrow 0$, so:

${\lim}_{x \rightarrow \infty} \frac{1}{{x}^{3} + 4} = \frac{0}{1 + 0}$
$\text{ } = 0$

We can verify this result by looking at the graph of $y = \frac{1}{{x}^{3} + 4}$

graph{ 1/(x^3+4) [-7, 13, -4.16, 5.84]}

and indeed it does appear that for large $x$ the function is approaching a horizontal asymptote $y = 0$