В однокруговом турнире по футболу любая команда набрала 10

First, we'll try to plug in the value:
lim_x to -oox+sqrt(x^2+2x) = -oo + sqrt(oo-oo)
We're already encountering a problem: it is simply not allowed to have oo-oo, it's like dividing by zero.
We need to try a different approach.
Whenever I see this kind of limit, I try to use a trick:
lim_x to -oox+sqrt(x^2+2x)
= lim_x to -oox+sqrt(x^2+2x)*(x-sqrt(x^2+2x))/(x-sqrt(x^2+2x))
These are the same becaus the factor we're multiplying with is essentially 1.
Why are we doing this? Because there exists a formula which says: (a-b)(a+b) = a^2-b^2
In this case a = x and b = sqrt(x^2+2x)
Let's apply this formula:
lim_x to -oo(x^2-(sqrt(x^2+2x))^2)/(x-sqrt(x^2+2x))
= lim_x to -oo(x^2-x^2-2x)/(x-sqrt(x^2+2x))
= lim_x to -oo(-2x)/(x-sqrt(x^2+2x))
Now we're going to use another trick. We'r going to use this one, because we want to get the x^2 out of the square root:
lim_x to -oo(-2x)/(x-sqrt(x^2(1+2/x))
If you look carefully, you see it's the same thing.
Now, you might say that sqrt(x^2) = x, but you have to remember that x is a negative number. Because we're taking the positive square root, sqrt(x^2) = -x in this case.
= lim_x to -oo(-2x)/(x+xsqrt(1+2/x))
= lim_x to -oo(-2x)/(x(1+sqrt(1+2/x)))
We can cancel the x:
= lim_x to -oo(-2)/(1+sqrt(1+2/x))
And now, we can finally plug in the value:
= -2/(1+sqrt(1+2/-oo))
A number divided by infinity, is always 0:
= -2/(1+sqrt(1+0)) = -2/(1+1) = -2/2 = -1
This is the final answer.
Hope it helps.