find the limit. (if the limit is infinite, enter ∞ or -∞, as ap\n lim _{x \rightarrow infty} \frac{sqrt{1+64…

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

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

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)