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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.
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.
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 $a=x^2+x$
HINT: i would write $$\frac{x\left(\sqrt{1+\frac{1}{x}}+1\right)}{x}$$ for $x$ tends to $\infty$
