Question

What is the limit as x approaches negative infinity of `x + sqrt(x^2 + 2x)`?

Answer 1

First, we'll try to plug in the value:
`lim_{x to -oo}x+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 -oo}x+sqrt(x^2+2x)`
`= lim_{x to -oo}x+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.