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{49x^{2}-40x}}{4x + 4+8x^{2}}

Explanation:

Step1: Divide numerator and denominator by $x^{2}$

For the numerator $\sqrt{49x^{2}-40x}$, we have $\frac{\sqrt{49x^{2}-40x}}{x^{2}}=\sqrt{\frac{49x^{2}-40x}{x^{4}}}=\sqrt{\frac{49}{x^{2}}-\frac{40}{x^{3}}}$. For the denominator $4x + 4+8x^{2}$, we have $\frac{4x + 4+8x^{2}}{x^{2}}=\frac{4}{x}+\frac{4}{x^{2}}+8$. So the limit becomes $\lim_{x
ightarrow\infty}\frac{\sqrt{\frac{49}{x^{2}}-\frac{40}{x^{3}}}}{\frac{4}{x}+\frac{4}{x^{2}}+8}$.

Step2: Use the limit properties

We know that $\lim_{x
ightarrow\infty}\frac{1}{x^{n}} = 0$ for $n>0$. So $\lim_{x
ightarrow\infty}\frac{49}{x^{2}}=0$, $\lim_{x
ightarrow\infty}\frac{40}{x^{3}}=0$, $\lim_{x
ightarrow\infty}\frac{4}{x}=0$ and $\lim_{x
ightarrow\infty}\frac{4}{x^{2}}=0$. Then $\lim_{x
ightarrow\infty}\frac{\sqrt{\frac{49}{x^{2}}-\frac{40}{x^{3}}}}{\frac{4}{x}+\frac{4}{x^{2}}+8}=\frac{\sqrt{0 - 0}}{0 + 0+8}$.

Answer:

$0$