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This question already has an answer here:

I've been struggling with the below problem (I'm a Calc I student and don't know how to approach this using the skills I've developed so far):

$ \lim _{x\to \infty }\left(\sqrt{x^2+4x}-x\right) $

I've tried rewriting using the conjugate:

$ \dfrac{4x}{\sqrt{x^2+4x}+x} $

But I'm not sure how to proceed. Writing the part of the denominator that is under the radical as a power of $ \frac{1}{2} $ is the only thing I can think of, but what emerges is really messy algebraically and I still don't know how to find the limit from there.

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marked as duplicate by Hans Lundmark, Joey Zou, Jack's wasted life, Joel Reyes Noche, R_D Sep 30 '16 at 6:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Hint: $$ \frac{4x}{\sqrt{x^2+4x}+x} =\frac{4x}{x\sqrt{1+4/x}+x}=\frac{4}{\sqrt{1+4/x}+1}$$

Note, that after taking $x$ out of the root we should have to use $|x|$ but as $x > 0$ it is the same as $x$.

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    $\begingroup$ +1 for mentioning the use of $|x|$! We tend to miss that fact sometimes! :) $\endgroup$ – FreezingFire Sep 29 '16 at 12:35
  • $\begingroup$ @FreezingFire That fact is superflous here. $ x \rightarrow \infty $ after all. $\endgroup$ – Dragonemperor42 Sep 29 '16 at 12:37
  • $\begingroup$ @Roby5 I know, but, say, if the limit was to $-\infty$? The limit would not be defined then (i.e. $-\infty$)! My point being, it is a good practice to factor it out of the radical as $|x|$ instead of just $x$. That is mentioned in this answer. $\endgroup$ – FreezingFire Sep 29 '16 at 12:45
  • $\begingroup$ @FreezingFire Maybe you are right. $\endgroup$ – Dragonemperor42 Sep 29 '16 at 12:46
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$$ \frac{4x}{\sqrt{x^2+4x}+x} =\frac{4x}{|x|\sqrt{1+4/x}+x}$$ Then $$\lim_{x \rightarrow +\infty} \frac{4x}{\sqrt{x^2+4x}+x}=\lim_{x \rightarrow +\infty}\frac{4x}{|x|\sqrt{1+4/x}+x}=\frac{4}{1+1}=2$$

$$\lim_{x \rightarrow -\infty} \frac{4x}{\sqrt{x^2+4x}+x}=\lim_{x \rightarrow -\infty}\frac{4x}{|x|\sqrt{1+4/x}+x}=\frac{4}{1-1}=+\infty$$

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  • $\begingroup$ For the second limit, it should be $-4/(1^+ -1)$ (because $x/|x|=-1$), so the limit tends to $-\infty$. $\endgroup$ – FreezingFire Sep 29 '16 at 12:48
  • $\begingroup$ No. For example. If $x=-10^6$, then $\left(\sqrt{x^2+4x}-x\right)=\left(\sqrt{10^{12}-4\cdot10^6}+10^6\right)$. Then $\left(\sqrt{x^2+4x}-x\right) \rightarrow +\infty$ $\endgroup$ – Roman83 Sep 29 '16 at 12:53
  • $\begingroup$ Sorry, my bad! In my comment it should have been $-4/(1^- - 1)$. You are correct. By the way, it is not my downvote! Your answer is perfectly correct. $\endgroup$ – FreezingFire Sep 29 '16 at 13:00
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Continuing from where you left

$$ \lim_{x \rightarrow \infty} \frac{4x}{\sqrt{x^2+4x}+x}=\lim_{x \rightarrow \infty} \frac{4}{\sqrt{1+\frac{4}{x}}+1}=2$$

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A good trick for solving limits when you get

$$\lim{f(x)}=\frac{'\infty'}{\infty}$$

Or

$$\lim{f(x)}=\frac{'0'}{0}$$

Is to divide the numerator and denominator by the highest power of $x$. In your case, the highest power is one so you divide top and bottom by $x$ (which is essentially just multiplying by 1).

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When $x$ tends to $+\infty$, simply use equivalents:

$x^2+4x\sim_{+\infty} x^2$, hence $\sqrt{x^2+4x}\sim_{+\infty} \sqrt{x^2}=x$, which means $$\sqrt{x^2+4x}=x+o(x),\quad\text{hence}\quad\sqrt{x^2+4x}+x=2x+o(x),$$so the fraction is $$ \frac{4x}{\sqrt{x^2+4x}+x}=\frac{4x}{2x+o(x)}=\frac 2{1+o(1)}\to 2. $$

When $x$ tends to $-\infty$, $\sqrt{x^2+4x}\sim_{-\infty} \sqrt{x^2}=\lvert x\rvert$, so (we may suppose $x<0$): $$\sqrt{x^2+4x}=\lvert x\rvert+o(\lvert x\rvert)-x=2\lvert x\rvert+o(\lvert x\rvert)\sim_{-\infty}2\lvert x\rvert\to+\infty.$$

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  • $\begingroup$ Actually, I had forgoten the 4 in the numerator. It's fixed now. Thanks for pointing the problem! $\endgroup$ – Bernard Sep 29 '16 at 12:40
  • $\begingroup$ $1/(2 + o(1))$ will necessarily tend to $1/2$. $o(1)$ tends to $0$, so the limit is indeed $1/2$. $\endgroup$ – Mariuslp Sep 29 '16 at 12:42
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    $\begingroup$ I think you're making a confusion between $O$ and $o$. $2$ is of $O(1)$, but not of $o(1)$. $\endgroup$ – Mariuslp Sep 29 '16 at 13:01
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    $\begingroup$ @MrYouMath: $o(1)$ simply means ‘tends to 0’. The (simplified) denominator tends to $1$. $\endgroup$ – Bernard Sep 29 '16 at 13:16
  • $\begingroup$ Sorry, I didn't know that there was a small o :D. Thank you. $\endgroup$ – MrYouMath Sep 29 '16 at 13:19

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