# How do you find lim sqrt(3x+1)-2sqrtx as x->oo?

Dec 12, 2017

${\lim}_{x \to \infty} \left(\sqrt{3 x + 1} - 2 \sqrt{x}\right) = - \infty$

#### Explanation:

The question asks to find ${\lim}_{x \to \infty} \left(\sqrt{3 x + 1} - 2 \sqrt{x}\right)$.

Multiply by $\frac{\sqrt{3 x + 1} + 2 \sqrt{x}}{\sqrt{3 x + 1} + 2 \sqrt{x}}$ to get
$= {\lim}_{x \to \infty} \left(\left(\sqrt{3 x + 1} - 2 \sqrt{x}\right) \cdot \frac{\sqrt{3 x + 1} + 2 \sqrt{x}}{\sqrt{3 x + 1} + 2 \sqrt{x}}\right)$
$= {\lim}_{x \to \infty} \left(\frac{3 x + 1 - 4 x}{\sqrt{3 x + 1} + 2 \sqrt{x}}\right)$
$= {\lim}_{x \to \infty} \left(\frac{1 - x}{\sqrt{3 x + 1} + 2 \sqrt{x}}\right)$
$= {\lim}_{x \to \infty} \left(\frac{1 - x}{\sqrt{x} \left(\sqrt{3 + \frac{1}{x}} + 2\right)}\right)$
$= {\lim}_{x \to \infty} \left(\frac{\frac{1}{\sqrt{x}} - \sqrt{x}}{\sqrt{3 + \frac{1}{x}} + 2}\right)$
$= \frac{{\lim}_{x \to \infty} \left(\frac{1}{\sqrt{x}} - \sqrt{x}\right)}{{\lim}_{x \to \infty} \left(\sqrt{3 + \frac{1}{x}} + 2\right)}$

The numerator evaluates to $- \infty$ while the denominator is a finite number.

Thus, the answer is $- \infty$.

This can be seen clearly with a graph:
graph{sqrt(3x+1)-2sqrt(x) [-1, 100, -5, 2]}