# How do you find lim sqrt(2x+3)-sqrtx as x->oo?

Jan 18, 2018

${\lim}_{x \rightarrow + \infty} \left(\sqrt{2 x + 3} - \sqrt{x}\right) = + \infty$

#### Explanation:

lim_(xrarr+oo)(sqrt(2x+3)-sqrtx)=?

$f \left(x\right) = \sqrt{2 x + 3} - \sqrt{x} =$

$\frac{\left(\sqrt{2 x + 3} - \sqrt{x}\right) \left(\sqrt{2 x + 3} + \sqrt{x}\right)}{\sqrt{2 x + 3} + \sqrt{x}}$ $=$

$\frac{{\sqrt{2 x + 3}}^{2} - {\sqrt{x}}^{2}}{\sqrt{2 x + 3} + \sqrt{x}}$ $=$

$\frac{2 x + 3 - x}{\sqrt{2 x + 3} + \sqrt{x}}$ $=$

$\frac{x + 3}{\sqrt{2 x + 3} + \sqrt{x}}$

As a result,

${\lim}_{x \rightarrow + \infty} \left(\sqrt{2 x + 3} - \sqrt{x}\right) = {\lim}_{x \rightarrow + \infty} \frac{x + 3}{\sqrt{2 x + 3} + \sqrt{x}}$ $=$

${\lim}_{x \rightarrow + \infty} \frac{x + 3}{\sqrt{{x}^{2} \left(\frac{2}{x} + \frac{3}{x} ^ 2\right)} + \sqrt{x}}$ $=$

${\lim}_{x \rightarrow + \infty} \frac{x + 3}{| x | \sqrt{\frac{2}{x} + \frac{3}{x} ^ 2} + | x | \sqrt{\frac{1}{x}}}$ $=$

$x \to + \infty$
$x > 0$

${\lim}_{x \rightarrow + \infty} \frac{x + 3}{x \sqrt{\frac{2}{x} + \frac{3}{x} ^ 2} + x \sqrt{\frac{1}{x}}}$ $=$

lim_(xrarr+oo)(cancel(x)(1+3/x))/(cancel(x)(sqrt(2/x+3/x^2)+sqrt(1/x)) $=$

lim_(xrarr+oo)(1+3/x)/(sqrt(2/x+3/x^2)+sqrt(1/x))=^((1/0^+) $+ \infty$