Limits with square roots

Question

I came across $\lim_{x\rightarrow\infty}\left(\sqrt{x^2+3}-x\right)=0$, but I don't quite understand how to calculate this. Could someone explain?

Answer 1

Hint:

$$\sqrt{x^2+3}-x={(\sqrt{x^2+3}-x)(\sqrt{x^2+3}+x)\over\sqrt{x^2+3}+x}$$

Can you take it from there?

Answer 2

$$ \sqrt{x^2+3} - x = \frac{(\sqrt{x^2+3} - x)(\sqrt{x^2+3} + x)} {\sqrt{x^2+3} + x} = \frac{3}{\sqrt{x^2+3} + x} $$ $${}$$ Evaluating something like $\displaystyle \lim_{x\to\infty} (\sqrt{x^2 + 8x} - x)$ would be more edifying because it shows that $\infty$ minus $\infty$ really is indeterminate. You get $$ \frac{8x}{\sqrt{x^2+8x} + x} = \frac 8 {\sqrt{1 + \frac 8 x} + 1} \to \frac 8 2 = 4 \text{ as } x\to+\infty. $$ And instead of $4$, you then see why any other number would have served as well.

Answer 3

Just another way to do it.

$$\sqrt{x^2+3} - x =x \left(\sqrt{1+\frac 3 {x^2}}-1\right)$$ Now, since $x$ is large, use the generalized binomial theorem or Taylor series $$\sqrt{1+y}=1+\frac{y}{2}+O\left(y^2\right)$$ Replace $y$ by $\frac 3 {x^2}$; this makes $$\sqrt{1+\frac 3 {x^2}}=1+\frac{3}{2 x^2}+O\left(\frac{1}{x^4}\right)$$ $$\sqrt{x^2+3} - x =x \left(1+\frac{3}{2 x^2}+O\left(\frac{1}{x^4}\right)-1\right)=\frac{3}{2 x}+O\left(\frac{1}{x^3}\right)$$

You could generalize the problem to $$\sqrt[k]{x^k+a} - x =x \left(\sqrt[k]{1+\frac a{x^k}}-1\right)$$ and use again the generalized binomial theorem or Taylor series $$\sqrt[k]{1+y}=1+\frac{y}{k}+O\left(y^2\right)$$ Replace $y$ by $\frac a {x^k}$; this makes $$\sqrt[k]{1+\frac a{x^k}}=1+\frac a {k x^k}+\cdots$$ and then $$\sqrt[k]{x^k+a} - x =\frac a {k x^{k-1}}+\cdots$$