Question

lhôpitals rule does not help with the limit below. find the limit some other way.

$$\lim_{x \to \infty} \frac{\sqrt{25x + 1}}{\sqrt{x + 2}}$$

$$\lim_{x \to \infty} \frac{\sqrt{25x + 1}}{\sqrt{x + 2}} = \square$$ (simplify your answer.)

Explanation:

Step1: Divide numerator and denominator by $\sqrt{x}$

$$\begin{align*} \lim_{x ightarrow\infty}\frac{\sqrt{25x + 1}}{\sqrt{x+2}}&=\lim_{x ightarrow\infty}\frac{\sqrt{\frac{25x + 1}{x}}}{\sqrt{\frac{x + 2}{x}}}\\ &=\lim_{x ightarrow\infty}\frac{\sqrt{25+\frac{1}{x}}}{\sqrt{1+\frac{2}{x}}} \end{align*}$$

Step2: Use the limit property $\lim_{x

ightarrow\infty}\frac{1}{x}=0$
As $x
ightarrow\infty$, $\frac{1}{x}
ightarrow0$ and $\frac{2}{x}
ightarrow0$.
Substitute these values into the expression:
$$\frac{\sqrt{25 + 0}}{\sqrt{1+0}}$$

Answer:

$5$