Question

question
determine the following limit in simplest form. if the limit is infinite, state that the limit does not exist (dne).
lim_{x
ightarrowinfty}\frac{sqrt{-21x^{3}+x^{6}}}{7 + 9x^{2}}
answer

Explanation:

Step1: Simplify the numerator

For large \(x\) values, when considering \(\sqrt{- 21x^{3}+x^{6}}\), since for \(x\to\infty\), the dominant term in \(-21x^{3}+x^{6}\) is \(x^{6}\). So \(\sqrt{-21x^{3}+x^{6}}=\sqrt{x^{6}(1 - \frac{21}{x^{3}})}=x^{3}\sqrt{1-\frac{21}{x^{3}}}\) (assuming \(x>0\) as \(x\to\infty\)).

Step2: Rewrite the limit

The original limit \(\lim_{x\to\infty}\frac{\sqrt{-21x^{3}+x^{6}}}{7 + 9x^{2}}\) becomes \(\lim_{x\to\infty}\frac{x^{3}\sqrt{1-\frac{21}{x^{3}}}}{7 + 9x^{2}}\).

Step3: Divide numerator and denominator by \(x^{2}\)

We get \(\lim_{x\to\infty}\frac{x\cdot\sqrt{1-\frac{21}{x^{3}}}}{\frac{7}{x^{2}}+9}\).

Step4: Evaluate the limit

As \(x\to\infty\), \(\frac{21}{x^{3}}\to0\) and \(\frac{7}{x^{2}}\to0\). So \(\lim_{x\to\infty}\frac{x\cdot\sqrt{1 - 0}}{0 + 9}=\lim_{x\to\infty}\frac{x}{9}=\infty\).

Answer:

The limit does not exist (DNE) as it is infinite.