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

[ronneedshelp]

I'm having a major problem with this one calculus problem. I multiplied by the conjugate and ended up with something that seemed workable, but then when I multiplied through by the reciprocal of the highest power of x in the denominator, the answer was 2/0. What do I do with that? Detailed explanation please . . . I have a quiz coming up soon and this one is stumping me.

My calculator is showing the graph tends towards -1. To get the answer via the methods we were taught in class, I multiplied the function by conjugate/conjugate (in this case, x - sqrt[x^2 + 2x] OVER x - sqrt[x^2 + 2x]), leaving me with -2x OVER (x - sqrt[x^2 + 2x])

I then multiplied this fraction by 1/x OVER 1/x (the highest power of x in the denominator), allowed all newly created fractions with x in the denominator to turn into zero, and was left with -2/(1-sqrt[1]) which is -2/0. I don't know what to do with this.

Anyone following my confusion here??
In response to Yaroslav . . . I'm not understanding what you mean by "substitute a negative x for positive x." Can you show me a step-by-step explanation???

[YaroslavY]

sfgrc and swimeasy are wrong - you have no right to either subtract negative infinities or take the square root of them. I'm sorry, I checked on the calculator, it is negative one.
You shouldn't have gotten -2/0 - you should've substituted -x, not positive x, since we are going to negative infinity. So you'll get 2/-2, which equals to negative one. It took me a while to figure this out

[calculus75]

its -1.

[kristabbbb]

sfgrc and swimeasy are wrong - you have no right to either subtract negative infinities or take the square root of them. I'm sorry, I checked on the calculator, it is negative one.
You shouldn't have gotten -2/0 - you should've substituted -x, not positive x, since we are going to negative infinity. So you'll get 2/-2, which equals to negative one. It took me a while to figure this out

hey can you help me with my math homework? im in calculus and i don't know how to solve:

find the value of the limit:

lim x approaches positive 0 while the equation is e^x squared - minus one, all divided by x

[Unregistered]

hahahahaha i love captchas

[Unregistered]

i am having a calculus midterm later, i am reviewing now, it comes out that i am having trouble with the squeeze theorem particularly this following exo : b) limx→∞
√
x sin
1
x
. (Hint: let y =
1
x
, and it becomes a one-sided
limit.)

can someone please help me with this????????
i would be most grateful ))))