Solving Limits Problems: limx->-inf & g(x)=x10/9

[Melawrghk]

Homework Statement

1. lim_x->-inf (sqrt(x²+6x-1) + x)
2. Let g(x) = x^10/9, find lim_h->0 g((20⁹+h) - g(20⁹))/h

Homework Equations

None that I know of.

The Attempt at a Solution

1. Well, first I put the -infinty in and I think it's an indetermination because it's inf-inf. So I decided to rationalize the equation and got:
6x-1
------------
sqrt(x²+6x-1) -x)
I figured the limit of the denominator has to be +infinity, limit of the top is -infinity, which would result in getting a -infinity. But the answer says it's -3.

2. I substituted and got this:
(20⁹+h)^10/9 - 20⁹
-----------------------
h

Which really gets me nowhere because it is still unclear at to what to do with the first part of the numerator. I'm only allowed to solve this with regular limit rules, nothign fancy...

 

[tiny-tim]

1. lim_x->-inf (sqrt(x²+6x-1) + x)
2. Let g(x) = x^10/9, find lim_h->0 g((20⁹+h) - g(20⁹))/h
…
I'm only allowed to solve this with regular limit rules, nothign fancy...

Hi Melawrghk!

(have an infinity: ∞ and a square-root: √ )

Hint: 1. can you solve lim_x->-∞ (√(x²+6x+9) + x)?

2. what is (1 + x)^10/9 ?

 

[Melawrghk]

Hint: 1. can you solve lim_x->-∞ (√(x²+6x+9) + x)?

Nope :) Because that is essentially what I'm asking. It turns out to be ∞-∞ and I'm not sure if the infinites are equal. Plus, it definitely doesn't equal -3...

2. what is (1 + x)^10/9 ?

Honestly, I have no clue. 9th root of (1+x)¹⁰?

Thanks for your help

 

[tiny-tim]

Nope :)

Yup

(what is √(x²+6x+9) ?)

Try again!

Honestly, I have no clue. 9th root of (1+x)¹⁰?

ok … try: what is (1 + x)¹⁰ if x is very small?

(1 + x)⁵ ?

(1 + x)^10/9 ?

 

[Dick]

Nope :) Because that is essentially what I'm asking. It turns out to be ∞-∞ and I'm not sure if the infinites are equal. Plus, it definitely doesn't equal -3...


Honestly, I have no clue. 9th root of (1+x)¹⁰?

Thanks for your help

For first one try this. sqrt(x^2+6x-1)=sqrt(x^2*(1+6/x-1/x^2))=|x|sqrt(1+6/x-1/x^2). Apply that to your rationalized form.

 

[Melawrghk]

Yup

(what is √(x²+6x+9) ?)

Try again!

OOH I'm dumb! :D Okay, I get it now. Can't believe I overlooked that...
EDIT: Wait, no. I get this one, but how would I do it with mine? Mine doesn't factor nicely...

ok … try: what is (1 + x)¹⁰ if x is very small?

(1 + x)⁵ ?

(1 + x)^10/9 ?

It's 1^whatever. But if I use that tactic, won't I just end up with 0/0 again? Because if I eliminate the "h" in the numerator, I'll be left with two equal but opposite terms... Sorry, I still don't get this one.

Thanks Dick, I'll try that.

 

[tiny-tim]

OOH I'm dumb! :D Okay, I get it now. Can't believe I overlooked that...
EDIT: Wait, no. I get this one, but how would I do it with mine? Mine doesn't factor nicely...

Use x² + 6x - 1 = (x² + 6x + 9)(1 - 10/(x² + 6x - 1))

It's 1^whatever. But if I use that tactic, won't I just end up with 0/0 again? Because if I eliminate the "h" in the numerator, I'll be left with two equal but opposite terms... Sorry, I still don't get this one.

(1 + x)¹⁰ = 1 +10x +15x² + …

so, for very small x, (1 + x)¹⁰ is approximately 1 +10x.

 

[Melawrghk]

Use x² + 6x - 1 = (x² + 6x + 9)(1 - 10/(x² + 6x - 1))

Where'd you get that?...

(1 + x)¹⁰ = 1 +10x +15x² + …

so, for very small x, (1 + x)¹⁰ is approximately 1 +10x.

Yeah.. binomial expansion. But that's for a nice power. 10/9 is ANYTHING but nice.
Maybe if I do the 9th root of 20⁹+h to the power of 10 that would help. I'll try.

PS. Thanks Dick, it worked :)