Evaluate the limit of function $\lim_{x\to\infty}\frac{(9x^2+1)^{1/2}}{x+2}.$

Question

Find the limit: $$\lim_{x\to\infty}\frac{(9x^2+1)^{1/2}}{x+2}.$$ I want to divide each of the terms by the highest power of $x$ but I failed to elimite the square root on it.

Answer 1

Hint:

\begin{align} \lim_{x\to\infty}\frac{(9x^2+1)^{1/2}}{x+2}&=\lim_{x\to\infty}\frac{\frac{\sqrt{9x^2+1}}{x}}{\frac{x+2}{x}}\\ &=\lim_{x\to\infty}\frac{\sqrt{\frac{9x^2+1}{x^2}}}{1+\frac{2}{x}}\\ &=\lim_{x\to\infty}\frac{\sqrt{9+\frac{1}{x^2}}}{1+\frac{2}{x}}\\ \end{align}

Answer 2

$$ \lim_{x\to\infty}\frac{(9x^2+1)^{1/2}}{x+2}\cdot\frac{1/x}{1/x} = \lim_{x\to\infty} \frac{(9+1/x^2)^{1/2}}{1+2/x}=3$$

Answer 3

HINT:

Set $\dfrac1x=h\implies h\to0^+, h>0$

$\sqrt{9x^2+1}=\sqrt{\dfrac{9+h^2}{h^2}}=\dfrac{\sqrt{9+h^2}}{|h|}=\dfrac{\sqrt{9+h^2}}h$ as $h>0$

Answer 4

Note that we can write

$$\begin{align} \frac{\sqrt{9x^2+1}}{x+2}&=\frac{3x}{x+2}\sqrt{1+\frac1{9x^2}}\\\\ &=3\left(1-\frac{2}{x+2}\right)\sqrt{1+\frac1{9x^2}} \end{align}$$

Since we have the limits

$$\lim_{x\to \infty}\left(1-\frac{2}{x+2}\right)=1$$

and

$$\lim_{x\to \infty}\sqrt{1+\frac1{9x^2}}=1$$

then the limit of interest is

$$\lim_{x\to \infty}\frac{\sqrt{9x^2+1}}{x+2}=3$$