Question

evaluate the limit (lim_{x
ightarrowinfty}\frac{sqrt{7 + 10x^{2}}}{10+6x})

Explanation:

Step1: Divide numerator and denominator by $x$

Divide each term in the numerator and denominator by the highest - power of $x$ in the denominator. Since the denominator is $\sqrt{7 + 10x^{2}}$, we divide by $x=\sqrt{x^{2}}$ (as $x\to\infty$, $x>0$).
\[

$$\begin{align*} \lim_{x\to\infty}\frac{\sqrt{7 + 10x^{2}}}{10+6x}&=\lim_{x\to\infty}\frac{\frac{\sqrt{7 + 10x^{2}}}{x}}{\frac{10 + 6x}{x}}\\ &=\lim_{x\to\infty}\frac{\sqrt{\frac{7}{x^{2}}+10}}{\frac{10}{x}+6} \end{align*}$$

\]

Step2: Evaluate the limit of each term

We know that $\lim_{x\to\infty}\frac{7}{x^{2}} = 0$ and $\lim_{x\to\infty}\frac{10}{x}=0$.
\[

$$\begin{align*} \lim_{x\to\infty}\frac{\sqrt{\frac{7}{x^{2}}+10}}{\frac{10}{x}+6}&=\frac{\sqrt{\lim_{x\to\infty}\frac{7}{x^{2}}+\lim_{x\to\infty}10}}{\lim_{x\to\infty}\frac{10}{x}+\lim_{x\to\infty}6}\\ &=\frac{\sqrt{0 + 10}}{0+6}\\ &=\frac{\sqrt{10}}{6} \end{align*}$$

\]

Answer:

$\frac{\sqrt{10}}{6}$