# Limit of square root without L'Hopital's rule.

How might one go about taking the following limit without using L'Hopital's rule? I am stumped:

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

• Intuitively it might be "close to but less than" (in some kind of handwaving sense) $$\lim_{x \to \infty} \sqrt{x^2 + x + \frac14} - x$$ – Henry Oct 28 '12 at 20:50
• @Henry's comment is very useful in fact, because it shows how it is possible to bound the function for positive $x$ - it is clearly greater than zero, and Henry puts it less than $\frac 1 2$ so the limit (if it exists) cannot be less than zero or greater than a half. – Mark Bennet Oct 28 '12 at 21:05
• I wonder how many times this has been answered already on this site... – Hans Lundmark Oct 28 '12 at 21:19
• – user17762 Oct 28 '12 at 22:16
• @HansLundmark I found the abstract duplicate version of this. Kindly vote to close this question as a duplicate of the above. – user17762 Oct 28 '12 at 22:17

$$\sqrt{x^2+x}-x=\sqrt{x^2\left(1+\frac1x\right)}-x=x\sqrt{1+\frac1x}-x=x\left(\sqrt{1+\frac1x}-1\right)\;;$$

let $u=\dfrac1x$, and this becomes $$\frac{\sqrt{1+u}-1}u=\frac{\sqrt{1+u}-1}u\cdot\frac{\sqrt{1+u}+1}{\sqrt{1+u}+1}=\frac{u}{u\left(\sqrt{1+u}+1\right)}=\frac1{\sqrt{1+u}+1}\;.$$

As $x\to\infty$, $u\to0^+$, so just calculate $$\lim_{u\to0^+}\frac1{\sqrt{1+u}+1}\;.$$

• I really like this approach. I wonder, if I couldn't use the power law either, how would I resolve $\lim_{x \to 0}\sqrt{1 + u}$? – providence Oct 29 '12 at 4:38
• @providence: You mean if you couldn’t appeal to the fact that $f(u)=\sqrt{1+u}$ is continuous at $0$ to say that its limit as $u\to0$ is $1$? You’d probably essentially have to prove it from the the definition of limit. – Brian M. Scott Oct 29 '12 at 4:41

HINT

$$\sqrt{x^2+x} - x = \dfrac{\left(\sqrt{x^2+x} - x\right) \left(\sqrt{x^2+x} + x \right)}{\sqrt{x^2+x} + x} = \dfrac{x^2+x - x^2}{\sqrt{x^2+x} + x}$$

Hint:

Multiply and divide function under limit by $\sqrt{x^2 + x} + x$: $\lim\limits_{x \to \infty}\left( \sqrt{x^2 + x} - x\right)=\lim\limits_{x \to \infty}\dfrac{\left( \sqrt{x^2 + x} - x\right)\left( \sqrt{x^2 + x} + x\right)}{ \sqrt{x^2 + x} + x}$

$$\lim_{x\to\infty} \sqrt{x^2+x}-x=\lim_{x\to\infty} (\sqrt{x^2+x}-x)\cdot \frac{ \sqrt{x^2+x}+x}{\sqrt{x^2+x}+x}=\lim_{x\to\infty} \frac{x^2+x-x^2}{\sqrt{x^2+x}+x}=\dots$$

Can you take it from here?

Observing that for $x>0$ we have $$\sqrt{x^2+x}-x=|x|\sqrt{1+\frac1x}-x=x\left(\sqrt{1+\frac1x}-1\right),$$ then using $(a-b)(a+b)=a^2-b^2$ gives us $$\left(\sqrt{x^2+x}-x\right)\cdot\left(\sqrt{1+\frac1x}+1\right)=x\left(1+\frac1x-1\right)=1.$$ Thus, $$\sqrt{x^2+x}-x=\frac1{\sqrt{1+\frac1x}+1},$$ which makes evaluation of the limit as $x\to\infty$ much simpler.

You could complete the square inside the square root to give:

$$\sqrt{\left(x+\frac 1 2\right)^2-\frac 1 4} - x=\left(x+\frac 1 2\right)\sqrt{1-\frac 1 {(2x+1)^2}}-x$$

And use the binomial series expansion of the square root. I mention this because it is a different idea from that in the other answers - it is not necessarily a better one. But if you are taking a factor outside the square root, I think completing the square gives you a better chance of getting the right factor. This puts a square in the denominator, rather than a linear denominator as in some of the other suggestions.

Hint: make a fraction of it and use $(a+b)(a-b)$.

Second hint: rewrite the fraction you've made (per Berci's comment) into $\frac{1}{\sqrt{1+\frac{1}{x}}+1}.$

First, factor out $x$ from the radical:

$$\lim_{x \to \infty} x \left(\sqrt{1+\frac{1}{x}} - 1 \right)$$

You can then Taylor expand the radical, as $1/x$ is small as $x$ becomes large. The relevant expansion is $(1+1/x)^{1/2} \approx 1 + \frac{1}{2x}$. This should make the result of the limit apparent.