Question

determine the following limit in simplest form. if the limit is infinite, state that the limit does not exist (dne).
\\(\lim_{x\to\infty}\frac{\sqrt{-37x^{2}+4x^{10}}}{6x^{2}+6 + 3x^{4}}\\)

Explanation:

Step1: Identify highest - power terms

For the numerator $\sqrt{-37x^{2}+4x^{10}}$, as $x\to\infty$, the highest - power term is $\sqrt{4x^{10}} = 2x^{5}$ (since for large $x$, $4x^{10}$ dominates $- 37x^{2}$). For the denominator $6x^{2}+6 + 3x^{4}$, the highest - power term is $3x^{4}$.

Step2: Divide numerator and denominator by $x^{4}$

We have $\lim_{x\to\infty}\frac{\frac{\sqrt{-37x^{2}+4x^{10}}}{x^{4}}}{\frac{6x^{2}+6 + 3x^{4}}{x^{4}}}=\lim_{x\to\infty}\frac{\sqrt{\frac{-37x^{2}}{x^{8}}+\frac{4x^{10}}{x^{8}}}}{\frac{6x^{2}}{x^{4}}+\frac{6}{x^{4}}+\frac{3x^{4}}{x^{4}}}=\lim_{x\to\infty}\frac{\sqrt{\frac{-37}{x^{6}} + 4x^{2}}}{\frac{6}{x^{2}}+\frac{6}{x^{4}}+3}$.

Step3: Evaluate the limit

As $x\to\infty$, $\frac{-37}{x^{6}}\to0$, $\frac{6}{x^{2}}\to0$, and $\frac{6}{x^{4}}\to0$. So the limit becomes $\lim_{x\to\infty}\frac{\sqrt{0 + 4x^{2}}}{0 + 0+3}=\lim_{x\to\infty}\frac{2x}{3}=\infty$.

Answer:

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