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Let's say we're trying to solve the following limit:

$\lim\limits_{x\to \infty} \sqrt{x^2+x}-x$

One way to do this is to use conjugates which results in the following:

$\lim\limits_{x\to \infty} \frac{\displaystyle x}{\displaystyle\sqrt{x^2+x}+x}$

Here's my problem: For the latter, we can simplify it so we'd have

$\lim\limits_{x\to \infty} \frac{\displaystyle x}{\displaystyle\sqrt{x^2+x} + x} = \lim\limits_{x\to \infty} \frac{\displaystyle x}{\displaystyle x+x} = \lim\limits_{x\to \infty} \frac{\displaystyle x}{\displaystyle 2x} = \frac{\displaystyle 1}{\displaystyle 2}$

but the same thing can't be done to the former. That is, we can't say

$\lim\limits_{x\to \infty} \sqrt{x^2+x}-x = \lim\limits_{x \to \infty} x - x = 0$

I believe this is issue comes from the last equality because $\lim\limits_{x \to \infty} x -x$ is like $\frac {0}{0}$; it's indeterminate but I'm not exactly sure.

EDIT

I'm so sorry I think I didn't phrase my question properly. My bad! I know why the two limits in the title are MUCH different; one tends to infinity whereas the other one has a finite value of $\frac{1}{2}$. I'm confused about why in the limit $\lim\limits_{x \to \infty}\frac{\displaystyle x}{\displaystyle \sqrt{x^2+x} + x}$, we can say the denominator is $2x$ but in the limit $\lim\limits_{x \to \infty} \sqrt{x^2+x} - x$ we can't say the square root is equal to $x$.

Thank you so much in advance!

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    $\begingroup$ I'm sorry to say this but nothing here is true. $\endgroup$
    – nonuser
    Commented Nov 9, 2019 at 17:06
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    $\begingroup$ @Aqua lol I was reading this thinking, either I'm crazy or has math just changed in the few years since I left school. $\endgroup$ Commented Nov 9, 2019 at 17:07
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    $\begingroup$ All your equations here are rather dubious IMHO. $\endgroup$ Commented Nov 9, 2019 at 17:07
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    $\begingroup$ @Aqua Other than the parts involving $x-x$, the rest is actually from a textbook and the answer is $\frac{1}{2}$ so I think the second and third limit are correct. $\endgroup$
    – user668217
    Commented Nov 9, 2019 at 17:09
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    $\begingroup$ It depends. Subtracting large, nearly identical quantities always leads to loss of accuracy. Something you may have seen with pocket calculators also :-) $\endgroup$ Commented Nov 9, 2019 at 17:56

2 Answers 2

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It is meaningless state that $\lim\limits_{x\to \infty} \sqrt{x^2+x}-x \ne x - x$ and $\lim\limits_{x\to \infty} \sqrt{x^2+x}+x = x + x$ we should state that

$$\lim\limits_{x\to \infty} \sqrt{x^2+x}-x = \lim\limits_{x\to \infty} \frac1{2}+o(1/x)=\frac12$$

and

$$\lim\limits_{x\to \infty} \sqrt{x^2+x}+x = \lim\limits_{x\to \infty} 2x+o(1)=\infty$$

the explanation in both case is in binomial first order approximation that is

$$\sqrt{x^2+x}=x\left(1+\frac1x\right)^\frac12= x\left(1+\frac1{2x}+o\left(\frac1x\right)\right)=x+ \frac1{2}+o\left(1\right) $$

which means that for $x$ large we have

$$\sqrt{x^2+x}\sim x+ \frac1{2}$$

and therefore

$$\sqrt{x^2+x}-x \sim \frac 12$$

$$\sqrt{x^2+x}+x \sim 2x+\frac 12$$


Edit

Note that for $\frac{x}{ \sqrt{x^2+x} + x}$ it is not correct to state that the denominator is $2x$ the complete steps are

$$\frac{x}{ \sqrt{x^2+x} + x}=\frac x x \frac{1}{ \sqrt{1+1/x} + 1} \to \frac12$$

For $\sqrt{x^2+x} - x$ indeed is not correct take$ \sqrt{x^2+x}=x$ what is true is that $\sqrt{x^2+x}\sim x+\frac12$.

If we use that approximation, it works with both limits. In this particular case first order approximation works and we can use it to evaluate both limits.

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  • $\begingroup$ I'm sorry but we haven't learned about the $o$ notation and binomial first order approximation. $\endgroup$
    – user668217
    Commented Nov 9, 2019 at 17:21
  • $\begingroup$ You can also use rationalization, the main issue doesn’t change, the statements in the title are meaningless. The two limit are simply different, one tends to 1/2 and the other to infinity. $\endgroup$
    – user
    Commented Nov 9, 2019 at 17:24
  • $\begingroup$ @BornaAhmadzade Even if you are not familiar with the little o notation, maybe you can appreciate the reason for which the two limit are so different. Indeed for $x$ large the $\sqrt{x^2+x}$ term behaves as $x+ \frac1{2}$ plus other terms which becomes negligible as $x$ becomes large and large. $\endgroup$
    – user
    Commented Nov 9, 2019 at 17:30
  • $\begingroup$ Ohhhh. I think I didn't phrase my question properly. My bad! I know why the two limits are MUCH different; one tends to infinity whereas the other one has a finite value of $\frac{1}{2}$. I'm confused about why in the limit $\lim\limits_{x \to \infty}\frac{\displaystyle x}{\displaystyle \sqrt{x^2+x} + x}$, we can say the denominator is $2x$ but in the limit $\lim\limits_{x \to \infty} \sqrt{x^2+x} - x$ we can't say the square root is equal to $x$. $\endgroup$
    – user668217
    Commented Nov 9, 2019 at 17:38
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    $\begingroup$ @BornaAhmadzade Form here $\frac{x}{ \sqrt{x^2+x} + x}$ it is not correct to state that the denominator is $2x$, the complete steps are $$\frac{x}{ \sqrt{x^2+x} + x}=\frac x x \frac{1}{ \sqrt{1+1/x} + 1} \to \frac12$$ $\endgroup$
    – user
    Commented Nov 9, 2019 at 17:44
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Hint: $$ (1\pm f(x))^n \approx 1 \pm nf(x) $$ for $f(x) \to 0$. The root from your question can be rewritten as: $$ \sqrt{x^2+x} = x\sqrt{1+\frac{1}{x}} \approx x(1+\frac{1}{2}\cdot\frac{1}{x}) $$ becasue from your limit $ 1/x \to 0 $. From here you can get the answers.

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  • $\begingroup$ Why the downvote :( ? $\endgroup$ Commented Nov 13, 2019 at 11:18

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