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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.

enter image description here

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.

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    $\begingroup$ In general, limits of polynomials can be treated as limits only of their highest degree terms, because those determine the limiting behavior. $\endgroup$ Oct 14, 2015 at 13:46
  • $\begingroup$ Where did you get this from? $\endgroup$
    – Soham
    Oct 14, 2015 at 13:47
  • $\begingroup$ The title should be changed though $\endgroup$
    – imranfat
    Oct 14, 2015 at 13:49
  • $\begingroup$ It's from a worksheet question from my Calculus class. $\endgroup$
    – Kelsey
    Oct 14, 2015 at 13:49

3 Answers 3

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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.

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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 $a=x^2+x$

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

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