I want to calculate the limit of $\lim_{x \to +\infty} (\sqrt{(x+a)(x+b)}-x)$

Question

I want to calculate the limit of: $$\lim_{x \to +\infty} (\sqrt{(x+a)(x+b)}-x) $$ $$(a,b ∈ R)$$

Now I know the result is $\frac{1}{2}(a+b)$, but I am having trouble getting to it.

Answer 1

Write $$\sqrt{(x+a)(x+b)}-x=\frac{(a+b)x+ab}{\sqrt{(x+a)(x+b)}+x}=\frac{a+b+\frac{ab}{x}}{\sqrt{(1+\frac ax)(1+\frac bx)}+1}.$$ Now let $x\to+\infty$.

Answer 2

A standard method is \begin{align} \lim_{x \to +\infty} (\sqrt{(x+a)(x+b)}-x) &= \lim_{x \to +\infty} (\sqrt{(x+a)(x+b)}-x) \frac{\sqrt{(x+a)(x+b)}+x}{\sqrt{(x+a)(x+b)}+x}\\ &= \lim_{x \to +\infty} \frac{(a+b)x+ab}{\sqrt{(x+a)(x+b)}+x}\\ &= \lim_{x \to +\infty} \frac{(a+b)+\frac{ab}{x}}{\sqrt{(1+\frac{a}{x})(1+\frac{b}{x})}+1}\\ \end{align}

A perhaps less standard method is to use the substitution $t=1/x$ so the limit becomes $$ \lim_{t\to0^+}\frac{\sqrt{(1+at)(1+bt)}-1}{t} $$ and you recognize the derivative at $0$ of the function $$ f(t)=\sqrt{(1+at)(1+bt)}=\sqrt{1+(a+b)t+abt^2} $$ Since $$ f'(t)=\frac{(a+b)+2abt}{2\sqrt{1+(a+b)t+abt^2}} $$ you're done.

Answer 3

Just another way to do it using Taylor expansions.

Since $x$ is large and positive $$A=\sqrt{(x+a)(x+b)}-x=\sqrt{x+a}\times\sqrt{x+b}-x=\sqrt x\times\sqrt{1+\frac a x}\times\sqrt x\times\sqrt{1+\frac b x}-x$$ $$A=x\left(\sqrt{1+\frac a x}\times\sqrt{1+\frac b x} -1\right)$$ Now, remembering that, for small values of $y$, $\sqrt {1+y} \approx 1+\frac y 2$, replace $y$ by $\frac a x$ in the first radical, then by $\frac b x$ in the second radical, expand and go to the limit.

Answer 4

Notice, $$\lim_{x\to \infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$ $$=\lim_{x\to +\infty}\frac{\left(\sqrt{(x+a)(x+b)}-x\right)\left(\sqrt{(x+a)(x+b)}+x\right)}{\left(\sqrt{(x+a)(x+b)}+x\right)}$$ $$=\lim_{x\to +\infty}\frac{x^2+(a+b)x+ab-x^2}{\left(\sqrt{(x+a)(x+b)}+x\right)}$$ $$=\lim_{x\to +\infty}\frac{(a+b)+\frac{ab}{x}}{\left(\sqrt{\left(1+\frac{a}{x}\right)\left(1+\frac{a}{x}\right)}+1\right)}$$

$$=\frac{a+b+0}{\left(\sqrt{\left(1+0\right)\left(1+0\right)}+1\right)}$$

$$=\frac{a+b}{1+1}=\frac{a+b}{2}$$

Answer 5

Let's call the expression $A$. First note that $$A=x((1+\frac{a+b}x+\frac{ab}{x^2})^{1/2}-1)$$ Use expansion with the little o notation $$A=x(1+\frac12(\frac{a+b}x+\frac{ab}{x^2}+o(\frac1x))+o(\frac1x)-1)=\frac12(a+b)+o(1)$$ as $x$ tends to infinity.