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

 A: 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:
$$
\lim_{n\to+\infty}f(x) = \lim_{n\to+\infty}(g(x) - x) = \lim_{n\to+\infty} \frac{(g(x))^n-x^n}{\sqrt[n]{(g(x))^{n-1}} +\sqrt[n]{(g(x))^{n-2}}x + \sqrt[n]{(g(x))^{n-3}}x^2 + \cdots + \sqrt[n]{g(x)}x^{n-2} +x^{n-1}} = \\
=\lim_{n\to+\infty} \frac{x^{n-1}a_1 + x^{n-1}a_2+ \cdots + x^{n-1}a_n + \cdots}{x^{n-1}\left(\sqrt[n]{1+o\left({1\over x}\right)+\cdots} + \sqrt[n]{1+o\left({1\over x}\right) +\cdots} +\cdots + 1\right)} = \\ 
= \frac{a_1 + a_2 + \cdots + a_n}{n}
$$
A: As an alternative, by $y=\frac1x \to 0$ and with $f(x)=\sqrt{\left(1+ay\right)\left(1+by\right)}$ we have
$$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)} - x\right) = \lim_{y\to0}\frac{\sqrt{\left(1+ay\right)\left(1+by\right)} - 1}{y}=f'(0)=\frac{a+b}2$$
A: $\small{(x+a)(x+b)=x^2+(a+b)x +ab=}$
$\small{(x+(a+b)/2)^2 +[ab-(a+b)^2/4]};$
$\small{C:=[ab-(a+b)^2/4]};$
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: $\small{(a+b)/2}$.
A: 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.
$$ \lim_{x\to \infty} \dfrac{(a+b) + \frac{ab}x}{\sqrt{1 + \frac{(a+b)}x + \frac{ab}{x^2}} + 1} = \dfrac{a+b}{2}$$
A: it is easily shown that as $z$ approaches zero 
$$
\log(1+z) = z + O(z^2)
$$
so we may write
$$
\log \sqrt{(x+a)(x+b)} =\log x +  \log \sqrt{(1+\frac{a}x)(1+\frac{b}x)} \\= \log x + \frac{a+b}{2x} + O(x^{-2})
$$
taking exponentials
$$
\sqrt{(x+a)(x+b)} = x\bigg(1+O(x^{-2})\bigg) \exp\frac{a+b}{2x} \\
= (x+O(x^{-1})\bigg(1+ \frac{a+b}{2x} + O(x^{-2}\bigg) \\
= x + \frac{a+b}2 +O(x^{-1})
$$
A: For a more informal answer...
$$\begin{align*}\sqrt{(x+a)(x+b)}-x&= \sqrt{x^2 + (a+b)x + ab} - x \\&= \sqrt{\left(\dfrac{a+b}{2} + x\right)^2-\frac{(a+b)^2}{4} +a b} - x \end{align*}$$
The expression $-\frac{1}{4} (-a-b)^2+a b$ is constant and is not important as $x \to \infty$ because the expression under the square root is dominated by the squared term.
Therefore
$$\begin{align*}\lim_{x\to\infty}\left(\sqrt{(x+a)(x+b)}-x\right)&= \\ \lim_{x\to\infty} \left(x + \dfrac{a+b}{2}\right) - x &= \dfrac{a+b}{2} \end{align*}$$
To add justification, consider $$\sqrt{ax^n + bx^m} = (ax^n + bx^m)^{1/2}, \quad n \geq m.$$
This is equal to $$(ax^n)^{1/2} + \frac{1}{2}\dfrac{bx^m}{\sqrt{ax^n}} - \dfrac{1}{8}\dfrac{b^2x^{2m}}{(\sqrt{ax^n})^3} ... $$
If we consider the case $n = 2, m = 1$ it is clear that the first two terms are non zero as $x \to \infty$ which is why we can't say that $\sqrt{x^2 - x} - x \approx 0 $ for large $x$.  On the other hand, the only term left in the case where $n = 2$ and $m = 0$ as $x \to \infty$ is the first term because every other term has a power of $x$ in the denominator.
A: There is another way to do it and even to get a bit more information.
Let
$$y=\sqrt{(x+a) (x+b)}-x=x \left(\sqrt{1+\frac{a}{x}}\times\sqrt{1+\frac{b}{x}} -1\right)$$ and now, remembering the binomial expansion or Taylor expansion
$$\sqrt{1+\epsilon}=1+\frac{\epsilon }{2}-\frac{\epsilon ^2}{8}+O\left(\epsilon ^3\right)$$ replace $\epsilon$ successively by $\frac{a}{x}$ and $\frac{b}{x}$ to get 
$$y=x \left(\left(1+\frac{a}{2 x}-\frac{a^2}{8 x^2}+O\left(\frac{1}{x^3}\right)\right)\times\left(1+\frac{b}{2 x}-\frac{b^2}{8 x^2}+O\left(\frac{1}{x^3}\right)\right) -1\right)$$ Expand the whole stuff to get
$$y=x \left(1+\frac{a+b}{2 x}-\frac{(a-b)^2}{8 x^2}+O\left(\frac{1}{x^3}\right)-1 \right)=\frac{a+b}{2}-\frac{(a-b)^2}{8 x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and also how it is approached.
