Values of the limit $\lim\limits_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$

Hi I have a question regarding finding the values of limit for the following question.

Let $a, b \in \mathbb R$. Find the limit

$$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$

When you have these kinds of limits it's always a good idea to multiply by conjugate. In your case you have: $$ \lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)} - x\right) = \lim_{x\to+\infty}\frac{\left(\sqrt{(x+a)(x+b)} - x\right)\left(\sqrt{(x+a)(x+b)} + x\right)}{\sqrt{(x+a)(x+b)} + x} = \\ = \lim_{x\to+\infty}\frac{(x+a)(x+b) - x^2}{\sqrt{(x+a)(x+b)} + x} = \lim_{x\to+\infty}\frac{x^2+ax+bx+ab-x^2}{\sqrt{(x+a)(x+b)} + x} = \\ = \lim_{x\to+\infty}\frac{x\left(a+b+\frac{ab}{x}\right)}{x\left(\sqrt{\left(1+{a\over x}\right)\left(1+{b\over x}\right)} + 1\right)} = \lim_{x\to+\infty}\frac{a+b+{ab\over x}}{\sqrt{1+{a+b\over x} + {ab \over x^2}}+1} $$

So as you see canceling $x$ from both nominator and denominator is going to produce the desired result.

In general consider functions in the form: $$ \begin{cases} f(x) = \sqrt[n]{(x + a_1)(x+a_2)\cdots (x+a_n)} - x \\ n,k \in \Bbb N \\ a_k \in \Bbb R \end{cases} $$

We know that: $$ a^n – b^n = (a – b)\left(a^{n – 1} + a^{n – 2}b + a^{n – 3}b^2 + \cdots + ab^{n – 2} + b^{n – 1}\right) $$

Denote: $$ g(x) = \sqrt[n]{(x + a_1)(x+a_2)\cdots (x+a_n)} $$ So based on that $f(x)$ may be rewritten as: $$ \lim_{n\to+\infty}f(x) = \lim_{n\to+\infty}(g(x) - x) = \lim_{n\to+\infty} \frac{(g(x))^n-x^n}{\sqrt[n]{(g(x))^{n-1}} +\sqrt[n]{(g(x))^{n-2}}x + \sqrt[n]{(g(x))^{n-3}}x^2 + \cdots + \sqrt[n]{g(x)}x^{n-2} +x^{n-1}} = \\ =\lim_{n\to+\infty} \frac{x^{n-1}a_1 + x^{n-1}a_2+ \cdots + x^{n-1}a_n + \cdots}{x^{n-1}\left(\sqrt[n]{1+o\left({1\over x}\right)+\cdots} + \sqrt[n]{1+o\left({1\over x}\right) +\cdots} +\cdots + 1\right)} = \\ = \frac{a_1 + a_2 + \cdots + a_n}{n} $$

As an alternative, by $y=\frac1x \to 0$ and with $f(x)=\sqrt{\left(1+ay\right)\left(1+by\right)}$ we have

$$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)} - x\right) = \lim_{y\to0}\frac{\sqrt{\left(1+ay\right)\left(1+by\right)} - 1}{y}=f'(0)=\frac{a+b}2$$

$\small{(x+a)(x+b)=x^2+(a+b)x +ab=}$

$\small{(x+(a+b)/2)^2 +[ab-(a+b)^2/4]};$

$\small{C:=[ab-(a+b)^2/4]};$

Let: $\small{y=(x+(a+b)/2)^2}$ then:

$\small{(y +C)^{1/2} - y^{1/2} +(a+b)/2}.$

Note that

$\small{ \lim_{y \rightarrow \infty}( (y+C)^{1/2} -y^{1/2})= 0}.$

Hence the limit is: $\small{(a+b)/2}$.

Hint: Multiply the numerator and denominator by the conjugate of the expression.

That is, you can try working with $$\dfrac{\sqrt{(x+a)(x+b)} - x}1\cdot\frac{\sqrt{(x + a)(x + b)} + x}{\sqrt{(x + a)(x + b)} + x} = \dfrac{(x+a)(x + b) - x^2}{\sqrt{(x + a)(x+b)} + x}$$

Expanding $(x + a)(x + b)$ in the numerator and simplifying gives us $$ \dfrac{(a+b)x + ab}{\sqrt{x^2 + (a+b)x + ab} + x}$$

Now divide numerator and denominator by $x$ to find your limit.

$$ \lim_{x\to \infty} \dfrac{(a+b) + \frac{ab}x}{\sqrt{1 + \frac{(a+b)}x + \frac{ab}{x^2}} + 1} = \dfrac{a+b}{2}$$