find the limit. (if the limit is infinite, enter ∞ or -∞, as ap
lim _{x
ightarrow infty} \frac{sqrt{1+64 x^{6}}}{3-x^{3}}
-8
excellent job!
resources

Question

find the limit. (if the limit is infinite, enter ∞ or -∞, as ap
 lim _{x
ightarrow infty} \frac{sqrt{1+64 x^{6}}}{3-x^{3}}
-8
excellent job!
resources

Accepted Answer

# Explanation:
## Step1: Divide numerator and denominator by $x^3$
When $x\to\infty$, we know that $\sqrt{x^{6}}=\vert x^{3}\vert=x^{3}$ (since $x\to\infty$ and $x > 0$).
$$
\begin{align*}
\lim_{x\to\infty}\frac{\sqrt{1 + 64x^{6}}}{3 - x^{3}}&=\lim_{x\to\infty}\frac{\frac{\sqrt{1+64x^{6}}}{x^{3}}}{\frac{3 - x^{3}}{x^{3}}}\
&=\lim_{x\to\infty}\frac{\sqrt{\frac{1}{x^{6}}+64}}{\frac{3}{x^{3}}-1}
\end{align*}
$$
## Step2: Use the limit property $\lim_{x\to\infty}\frac{1}{x^{n}} = 0$ ($n>0$)
We know that $\lim_{x\to\infty}\frac{1}{x^{6}}=0$ and $\lim_{x\to\infty}\frac{3}{x^{3}}=0$.
$$
\begin{align*}
\lim_{x\to\infty}\frac{\sqrt{\frac{1}{x^{6}}+64}}{\frac{3}{x^{3}}-1}&=\frac{\sqrt{\lim_{x\to\infty}\frac{1}{x^{6}}+64}}{\lim_{x\to\infty}\frac{3}{x^{3}}-1}\
&=\frac{\sqrt{0 + 64}}{0-1}\
&=\frac{8}{-1}=- 8
\end{align*}
$$

# Answer:
$-8$

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QUESTION IMAGE

Question

Explanation:

Step1: Divide numerator and denominator by \(x^3\)

Step2: Use the limit property \(\lim_{x\to\infty}\frac{1}{x^{n}} = 0\) (\(n>0\))

Answer: