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Determine the limit of the following or prove it doesn't exist:
$$ lim_{xto + infty}left(sqrt{(x+a)(x+b)} - xright) space space text{where} space a,b in mathbb R$$

I know that the limit is supposed to equal $frac {a+b}{2}$ but frankly have no idea how to get there.

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

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

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

$$lim_{xto+infty}left(sqrt{(x+a)(x+b)} - xright) = lim_{yto0}frac{sqrt{left(1+ayright)left(1+byright)} - 1}{y}=f'(0)=frac{a+b}2$$

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

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

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

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: $small{(a+b)/2}$.

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
$$
sqrt{(x+a)(x+b)} = xbigg(1+O(x^{-2})bigg) expfrac{a+b}{2x} \
= (x+O(x^{-1})bigg(1+ frac{a+b}{2x} + O(x^{-2}bigg) \
= x + frac{a+b}2 +O(x^{-1})
$$

There is another way to do it and even to get a bit more information.

Let
$$y=sqrt{(x+a) (x+b)}-x=x left(sqrt{1+frac{a}{x}}timessqrt{1+frac{b}{x}} -1right)$$ and now, remembering the binomial expansion or Taylor expansion
$$sqrt{1+epsilon}=1+frac{epsilon }{2}-frac{epsilon ^2}{8}+Oleft(epsilon ^3right)$$ replace $epsilon$ successively by $frac{a}{x}$ and $frac{b}{x}$ to get
$$y=x left(left(1+frac{a}{2 x}-frac{a^2}{8 x^2}+Oleft(frac{1}{x^3}right)right)timesleft(1+frac{b}{2 x}-frac{b^2}{8 x^2}+Oleft(frac{1}{x^3}right)right) -1right)$$ Expand the whole stuff to get
$$y=x left(1+frac{a+b}{2 x}-frac{(a-b)^2}{8 x^2}+Oleft(frac{1}{x^3}right)-1 right)=frac{a+b}{2}-frac{(a-b)^2}{8 x}+Oleft(frac{1}{x^2}right)$$ which shows the limit and also how it is approached.