How Does the ($\sqrt{x^2+x}+x)$ Equal $(\sqrt{x^2}+x)$ When Calculating The Limit of Infinity?

Question

I am asking this because of the following question:

What is the Limit of positive infinity for the equation $\frac{1}{\sqrt{x^2+x}+x}$?

The following steps are done to get the answer, which is 2.

I am not sure how the 3rd step went from having a numerator of $\sqrt{x^2+x}+x$ to having a numerator of $\sqrt{x^2}+x$. It's as if the $\sqrt{x}$ just disappeared.

Can anyone explain why this happens?

All help is appreciated.

Answer 1

While the result of the calculation is correct, it is not entirely trivial to justify the step you are having trouble with using "standard" theorems so I would suggest to keep manipulating the expression in the end of the first line algebraically as in

$$ \frac{\sqrt{x^2 + x} + x}{x} = \frac{\sqrt{x^2 + x}}{\sqrt{x^2}} + 1 = \sqrt{1 + \frac{1}{x}} + 1$$

resulting in an expression that clearly tends to $2$ as $x \to \infty$. This manipulation is valid as $x \to \infty$ and so we can assume that $x$ is positive and so $\sqrt{x^2} = |x| = x$. This looks much less "magical" and will cause you to make less mistakes and also justifies why the $x$ term inside the root (which gives rise to the $\frac{1}{x}$ inside the root) can be ignored in this case in for the calculation of the limit.

Answer 2

Difference of two squares: $(a-b)(a+b)=a^2-b^2$

Or in this case $(\sqrt{a}-x)(\sqrt{a}+x)=a-x^2$ where $a=x^2+x$

Answer 3

HINT: i would write $$\frac{x\left(\sqrt{1+\frac{1}{x}}+1\right)}{x}$$ for $x$ tends to $\infty$