# How do you find the limit of  (sqrt(x^2+10x+1)-x) as x approaches oo?

Sep 3, 2016

${\lim}_{x \to \infty} \left(\sqrt{{x}^{2} + 10 x + 1} - x\right) = 5$

#### Explanation:

Let $c$ be a constant.

Then we find:

${\lim}_{t \to \infty} \left(\sqrt{{t}^{2} + c} - t\right) = {\lim}_{t \to \infty} \left(\sqrt{{t}^{2} + c} - t\right)$

$\textcolor{w h i t e}{{\lim}_{t \to \infty} \left(\sqrt{{t}^{2} + c} - t\right)} = {\lim}_{t \to \infty} \frac{\left(\sqrt{{t}^{2} + c} - t\right) \left(\sqrt{{t}^{2} + c} + t\right)}{\sqrt{{t}^{2} + c} + t}$

$\textcolor{w h i t e}{{\lim}_{t \to \infty} \left(\sqrt{{t}^{2} + c} - t\right)} = {\lim}_{t \to \infty} \frac{\left({t}^{2} + c\right) - {t}^{2}}{\sqrt{{t}^{2} + c} + t}$

$\textcolor{w h i t e}{{\lim}_{t \to \infty} \left(\sqrt{{t}^{2} + c} - t\right)} = {\lim}_{t \to \infty} \frac{c}{\sqrt{{t}^{2} + c} + t}$

$\textcolor{w h i t e}{{\lim}_{t \to \infty} \left(\sqrt{{t}^{2} + c} - t\right)} = 0$

$\textcolor{w h i t e}{}$
Note that ${x}^{2} + 10 x + 1 = {\left(x + 5\right)}^{2} - 24$

Let $t = x + 5$ and $c = - 24$

Then:

${\lim}_{x \to \infty} \left(\sqrt{{x}^{2} + 10 x + 1} - x\right) = {\lim}_{t \to \infty} \left(\sqrt{{t}^{2} + c} - t + 5\right)$

$\textcolor{w h i t e}{{\lim}_{x \to \infty} \left(\sqrt{{x}^{2} + 10 x + 1} - x\right)} = {\lim}_{t \to \infty} \left(\sqrt{{t}^{2} + c} - t\right) + 5$

$\textcolor{w h i t e}{{\lim}_{x \to \infty} \left(\sqrt{{x}^{2} + 10 x + 1} - x\right)} = 0 + 5$

$\textcolor{w h i t e}{{\lim}_{x \to \infty} \left(\sqrt{{x}^{2} + 10 x + 1} - x\right)} = 5$