# Why is $\lim\limits_{x\to \infty} \sqrt{x^2+x}-x \ne x - x$ but $\lim\limits_{x\to \infty} \sqrt{x^2+x}+x = x + x$?

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:

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

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

$$\lim\limits_{x\to \infty} \fracx}{\displaystyle\sqrt{x^2+x} + x} = \lim\limits_{x\to \infty} \fracx}x+x} = \lim\limits_{x\to \infty} \fracx}2x} = \frac1}2}$$

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 $$\lim\limits_{x \to \infty}\fracx}\sqrt{x^2+x} + x}$$, 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!

• I'm sorry to say this but nothing here is true. Nov 9, 2019 at 17:06
• @Aqua lol I was reading this thinking, either I'm crazy or has math just changed in the few years since I left school. Nov 9, 2019 at 17:07
• All your equations here are rather dubious IMHO. Nov 9, 2019 at 17:07
• @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.
– user668217
Nov 9, 2019 at 17:09
• It depends. Subtracting large, nearly identical quantities always leads to loss of accuracy. Something you may have seen with pocket calculators also :-) Nov 9, 2019 at 17:56

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.

• I'm sorry but we haven't learned about the $o$ notation and binomial first order approximation.
– user668217
Nov 9, 2019 at 17:21
• 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.
– user
Nov 9, 2019 at 17:24
• @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.
– user
Nov 9, 2019 at 17:30
• 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 $\lim\limits_{x \to \infty}\fracx}\sqrt{x^2+x} + x$, 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 $x$.
– user668217
Nov 9, 2019 at 17:38
• @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 $$\frac{x}{ \sqrt{x^2+x} + x}=\frac x x \frac{1}{ \sqrt{1+1/x} + 1} \to \frac12$$
– user
Nov 9, 2019 at 17:44

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.

• Why the downvote :( ? Nov 13, 2019 at 11:18