find the limit. (if the limit is infinite, enter ∞ or -∞, as appropriate. if the limit does not otherwise exist, enter dne.)
limₓ→∞ x³ / √(x⁶ + 7)

Question

Accepted Answer

# Explanation:
## Step1: Divide numerator and denominator by $ x^3 $
Since $ x\to\infty$, we know that $ x^3=\sqrt{x^6}$.
$$
\begin{align*}
\lim_{x\to\infty}\frac{x^3}{\sqrt{x^6 + 7}}&=\lim_{x\to\infty}\frac{x^3/x^3}{\sqrt{x^6+7}/x^3}\
&=\lim_{x\to\infty}\frac{1}{\sqrt{\frac{x^6 + 7}{x^6}}}\
\end{align*}
$$
## Step2: Simplify the expression inside the square - root
Using the property $ \frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}$, we have $\frac{x^6+7}{x^6}=1+\frac{7}{x^6}$.
$$
\begin{align*}
\lim_{x\to\infty}\frac{1}{\sqrt{\frac{x^6 + 7}{x^6}}}&=\lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{7}{x^6}}}\
\end{align*}
$$
## Step3: Use the limit property $\lim_{x\to\infty}\frac{1}{x^n}=0$ for $n>0$
As $x\to\infty$, $\lim_{x\to\infty}\frac{7}{x^6}=0$. Then $\lim_{x\to\infty}\sqrt{1+\frac{7}{x^6}}=\sqrt{1 + 0}=1$.
$$
\begin{align*}
\lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{7}{x^6}}}&=\frac{1}{1}\
\end{align*}
$$

# Answer:
$1$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

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

Step2: Simplify the expression inside the square - root

Step3: Use the limit property \(\lim_{x\to\infty}\frac{1}{x^n}=0\) for \(n>0\)

Answer: