Question

find the limit. (if the limit is infinite, enter ∞ or -∞, as appropriate. if the limit does not otherwise exist, enter dne.
\lim_{x \to -\infty} \frac{\sqrt{1 + 64x^6}}{5 - x^3}

Explanation:

Step1: Analyze the highest degree terms

For the numerator $\sqrt{1 + 64x^6}$, the highest degree term inside the square root is $64x^6$, so $\sqrt{1 + 64x^6}\approx\sqrt{64x^6}=|8x^3|$. As $x
ightarrow -\infty$, $x^3<0$, so $|8x^3|=-8x^3$.
For the denominator $5 - x^3$, the highest degree term is $-x^3$.

Step2: Simplify the limit

Substitute the approximations into the limit:

$$ \lim_{x ightarrow -\infty}\frac{\sqrt{1 + 64x^6}}{5 - x^3}=\lim_{x ightarrow -\infty}\frac{-8x^3}{-x^3} $$

Step3: Cancel out common terms

Cancel out $x^3$ (note that $x
eq0$ as $x
ightarrow -\infty$):

$$ \lim_{x ightarrow -\infty}\frac{-8x^3}{-x^3}=\lim_{x ightarrow -\infty}8 $$

Step4: Evaluate the limit

The limit of a constant is the constant itself, so $\lim_{x
ightarrow -\infty}8 = 8$.

Answer:

8