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

Question

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!

Answer 1

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.

Answer 2

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.