Can you please help with a limit problem in calculus.?

Question

The limit as x→-∞ of (x + the square root of (x^2 + 2x)). I have to solve it algebraically. I keep getting DNE but when I graph it, the answer looks like it should be -1. Can you please show detailed work on how to do this problem or a similar one. Here's what I've tried but obviously something is wrong: *sq stands for square root of---
x + sq(x^2 + 2x) * (x – sq(x^2 +2x))
--------------------------------------------
1* ((x – sq(x^2 +2x))

x^2 – (x^2+2x)
------------------
x – sq(x^2 +2x)

-2x / x
-----------
(x-sq(x^2 + 2x))/x

-2
-----
1-sq(1-(2x^-.5))

I took the limits of each part to get:

-2
----
1-sq(1-0)

-2
----
0

DNE

It might be easiest to write this out on paper first because I couldn't type alll the mathamatical symbles -Thanks

thanks Pascal----that makes sense but why don't you devide the 1 that isn't under the square root by -sq(x^2) too. That would make the denominator -1 + sq(1+2/x). Sorry....I'm still not getting that part

Answer 1

Here's what's wrong:

-2x / x
-----------
(x-sq(x^2 + 2x))/x

-2
-----
1-sq(1-(2x^-.5))


You divided the numerator correctly, but the denominator is screwed up. Let's just focus on the square root:

-√(x²+2x)/x
In order to divide this, we have to put x under a square root sign. Thus we have -√(x²+2x)/-√(x²). Please note: negative square root of x². x² has two roots, and since we are taking x→-∞, we are dividing by the negative one, not the positive one. We can now apply the rule √a/√b=√(a/b) to get √(1+2/x) (not (1-2/√x), as you incorrectly put in your original problem). Thus the correct result at this point is:

[x→-∞]lim -2/(1+√(1+2/x))
-2/2
-1

Which agrees with the graph. The main problem in your derivation is that you incorrectly took -√(x²+2x)/x as being -√(x²+2x)/√(x²) instead of -√(x²+2x)/-√(x²), and that sign error caused you to get 0 for the denominator instead of the correct result of 2.

Edit: "that makes sense but why don't you devide the 1 that isn't under the square root by -sq(x^2) too. That would make the denominator -1 + sq(1+2/x). Sorry....I'm still not getting that part"

That's easy to answer. I _am_ dividing it by -√(x²). Remember x=-√(x²), so x/x = x/-√(x²) = -√(x²)/-√(x²) = 1. What you can't do, is take x/x = x/-√(x²) = x/-x = -1, or x/x = x/-√(x²) = √(x²)/-√(x²) = -1, because x ≠ √(x²), rather x = -√(x²). I think you have a tendency to see -√(x²) as being equal to -x, rather than x, because they both have negative signs in front of them. This is a mistake, albeit an easy one to make. You must recognize that -x, despite being written with a negative sign, is a _positive_ number. Conversely, -√(x²) is a negative number, as is x. Then it will be easier to see why we replace x with -√(x²), and why we must do this for all instances of x, not just the ones we're dividing by.

That help?

Answer 2

The problem with your solution is that there are two square roots of every #.
Try this way:

make the change of variable x=y-1
then as x→-∞ , y→-∞
and the equation becomes:

y-1 + √((y-1)^2 + 2(y-1)) = y - 1 + √(y^2 - 2y + 1 + 2y - 2)

= y - 1 + √(y^2 - 1)

and since limit as y→-∞ of √(y^2 - 1) = -y,

[√(y^2 - 1) = -y
divide by -y
√(1 - 1/y^2) = 1
taking limit as y→-∞
√(1 - 0) = 1
1 = 1
equation checks]

therefore, lim y→-∞ of y - 1 + √(y^2 - 1) = -1