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:
$$
\lim_{n\to+\infty}f(x) = \lim_{n\to+\infty}(g(x) - x) = \lim_{n\to+\infty} \frac{(g(x))^n-x^n}{\sqrt[n]{(g(x))^{n-1}} +\sqrt[n]{(g(x))^{n-2}}x + \sqrt[n]{(g(x))^{n-3}}x^2 + \cdots + \sqrt[n]{g(x)}x^{n-2} +x^{n-1}} = \\
=\lim_{n\to+\infty} \frac{x^{n-1}a_1 + x^{n-1}a_2+ \cdots + x^{n-1}a_n + \cdots}{x^{n-1}\left(\sqrt[n]{1+o\left({1\over x}\right)+\cdots} + \sqrt[n]{1+o\left({1\over x}\right) +\cdots} +\cdots + 1\right)} = \\ 
= \frac{a_1 + a_2 + \cdots + a_n}{n}
$$
