# How to compute $\lim_{x\to \infty} x \sqrt{x^2 + 1} - x$?

I'm not sure which strategy I should use. The indeterminate form is $$\infty - \infty$$, but I have not found a way to apply l'Hospital yet. Intuitively, this limit seems to diverge because the $$x\sqrt{x^2 + 1} \approx x^2$$ and $$x^2$$ grows a lot faster than $$x$$. In other words, the left term dominates the right one. So I'm looking for a direction to prove the fact. It doesn't have to be l'Hospital.

• Hint: factor out an $x$.
– dxiv
Jun 18 at 1:28
• All you need is $\sqrt{a+1} \geq \sqrt{a}$. Can you prove that? Jun 18 at 1:37
• I think I can prove that. The derivative of $\sqrt{x}$ is always positive, so it's an increasing function. Therefore, $\sqrt{x + 1| \geq \sqr{x}$. But I don't clearly see how that lets me to solve the problem. Jun 18 at 15:28
• For $x\ge\frac34$, we have $\sqrt{x^2+1}\ge\frac54$, therefore, $x\sqrt{x^2+1}-x\ge\frac x4$. A more interesting question would be $\lim\limits_{x\to\infty}\left(x\sqrt{x^2+1}-x^2\right)$.
– robjohn
Jun 18 at 17:51

Hint: Multiply with the quadratic conjugate i.e. multiply with $$\frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}$$

• Thank you! I did it. First I factored $x$ out. Then I multiplied by your quadratic conjugate. I got $x^3/(\sqrt{x^2 +1} + 1)$, which takes the form $\infty/\infty$. Then I applied l'Hospital twice getting $6x\sqrt{x^2+1}/2 + 3x^3/2\sqrt{x^2+1}$, whose term on the right is still unclear but also taking the form $\infty/\infty$, so applying l'Hospital on it alone yields $9x^2\sqrt{x^2 +1}/2$ which goes to $\infty$. So l'Hospital alone and your quadratic conjugate prove this limit diverges. Thank you so much. Jun 18 at 15:38
• Oops. It's much easier than that. I made mistakes above for sure. Here it is again. After multiplying by the quadratic conjugate, we get $x^3/(\sqrt{x^2 +1} + 1)$. Applying l'Hospital, we end up with $3x\sqrt{x^2 + 1}$, which explodes up. Jun 18 at 16:11

My approach would be like this:

$$x\sqrt{x^2+1}-x=x(\sqrt{x^2+1}-1)\geq 2x$$ when $$x\geq 2\sqrt 2$$. And when $$x\to +\infty$$, $$2x\to +\infty$$, hence $$x\sqrt{x^2+1}-x\to +\infty$$.

The question is kind of strange though because at first glance I thought it was $$\sqrt{x^2+1}-x$$. It would be a little more interesting: $$\sqrt{x^2+1}-x=x(\sqrt{1+\frac{1}{x^2}}-1)=x(1+\frac{1}{2}\frac{1}{x^2}+o(\frac{1}{x^2})-1)=x(\frac{1}{2x^2}+o(\frac{1}{x^2}))\to 0(x\to +\infty)$$.

The expression $$x\sqrt{x^2+1}-x$$ can be rewritten, by factoring $$x$$, as $$x(\sqrt{x^2+1}-1)$$.

As $$x \rightarrow \infty$$, $$\sqrt{x^2+1} \rightarrow x$$ (informally, think that the $$1$$ will not affect on the large value inside the radical).

Therefore, the expression $$x(\sqrt{x^2+1}-1)$$ can be written as $$x(x-1)$$ [valid for $$x \rightarrow \infty$$].

As $$x \rightarrow \infty$$, $$x-1 \rightarrow x$$ (informally, think that subtracting $$1$$ will not affect on large value of $$x$$).

Therefore, $$x(x-1)$$ can be written as $$x \cdot x = x^2$$ [valid for $$x \rightarrow \infty$$].

Clearly, $$x^2$$ tends to infinity as $$x$$ tends to infinity.

$$\implies \lim_{x \rightarrow \infty} \bigg( x\sqrt{x^2+1}-x \bigg) \rightarrow \infty$$

• It's great that you know how use the written language to show your thought process. So much easier to read (and pretty to see). Thank you so much. Jun 18 at 15:42

For any $$x>0$$, we have $$x \sqrt{x^2+1}-x \geqslant x \sqrt{x^2}-x=x^2(x-1) \rightarrow \infty$$ Therefore $$\textrm{ as } x\rightarrow \infty ,$$

$$\qquad\qquad x \sqrt{x^2+1}-x \rightarrow \infty .$$

• Very nice argument. I begin to realize what Brian Moehring had meant. Jun 18 at 15:39

Simply,

$$\sqrt{1 + x^2} > \sqrt{x^2} = |x| = x\,, \quad x > 0 \\ \therefore x\sqrt{1 + x^2} > x^2\,, \quad x > 0 \\ x\sqrt{1 + x^2} - x > x^2 - x\,, \quad x > 0$$

$$\lim\limits_{x \to \infty}{x^2 - x} = \lim\limits_{x \to \infty}{\left( x - \dfrac{1}{2}\right)^2 - \dfrac{1}{4}} \approx \lim\limits_{x \to \infty}{x^2} \to \infty$$

$$\lim\limits_{x \to \infty}{x\sqrt{1 + x^2} - x} \to \infty$$

Alternatively, one could use logarithms, since $$f(x) > \ln{f(x)}$$

$$\sqrt{1 + x^2} > \sqrt{x^2} = |x| = x\,, \quad x > 0 \\\sqrt{1 + x^2} - 1 > \ln{\left(\sqrt{1 + x^2} - 1\right)} > \ln{(x - 1)}$$

Now, $$\lim\limits_{x \to \infty}{\ln{(x - 1)}} \to \infty$$. The rest should be clear.

As suggested in the comments, we have that

$$x \sqrt{x^2 + 1} - x =x\cdot \left(\sqrt{x^2 + 1} - 1\right) \to \infty \cdot \infty=\infty$$