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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Find the limit without using L'hopitals rule.

limit of sqrt (x^2 + x + 1) + x as x approaches negative infinity.

I've been playing with it for about an hour to no avail. :-(

A logarithm is just a misspelled algorithm.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

I guess the limit is negative one-half, but I may not be exactly right.

My logic is that 7.5 x 7.5 is 56.25 and 7x8 is 56.

**igloo** **myrtilles** **fourmis**

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

The answer is 1/2 or -1/2, I don't remember. I'd just like to know how did they find it.

A logarithm is just a misspelled algorithm.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

-1/2, although I'm as stumped as you are.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**numen****Member**- Registered: 2006-05-03
- Posts: 115

I don't know what L'hopitals rule is, but here's what I did (hoping that I didn't use it!):

Bang postponed. Not big enough. Reboot.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

If you don't know L'H rule, then you can be sure that you didn't use it. L'H rule is:

If f(x) goes to 0 and g(x) goes to 0 as x goes to c, then:

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**numen****Member**- Registered: 2006-05-03
- Posts: 115

It's not even in my calculus textbook.

Bang postponed. Not big enough. Reboot.

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**Ricky****Moderator**- Registered: 2005-12-04
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I can just about guarantee you that it is. The equivalent would be writing a calculus text book and forgetting to talk about Riemann sums.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Ricky, doesn't L'hopitals rule only apply if it is in an indeterminant form such as 0/0 or infinity/infinity?

A logarithm is just a misspelled algorithm.

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**Ricky****Moderator**- Registered: 2005-12-04
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Yes, infinity/infinity works as well.

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**mikau****Member**- Registered: 2005-08-22
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yes but I thought it ONLY worked in that case. Not true?

A logarithm is just a misspelled algorithm.

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**Ricky****Moderator**- Registered: 2005-12-04
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I'm starting to get the feeling you missed the line that I wrote:

If f(x) goes to 0 and g(x) goes to 0 as x goes to c, then:

But yes, it must be that f(x) and g(x) both go to 0 or +/- infinity.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

ok, right.

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

If you don't know L'H rule, then you can be sure that you didn't use it.

Not true! I think what he meant was he's not familiar what "L'hopitals rule" refers to so he wouldn't know if he was using it. Just like a lot of people forget the transiative and associative property are, but they use them every day. You know the property, but forget the name.

Anyway, thanks Numen for the solution. you often forget you can factor out an x in an expression like x + 1. That also causes the appearance of terms like 1/x^n which are valuable in limit problems. Thanks again!

*Last edited by mikau (2006-09-15 07:56:34)*

A logarithm is just a misspelled algorithm.

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**numen****Member**- Registered: 2006-05-03
- Posts: 115

No problem : )

Bang postponed. Not big enough. Reboot.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Right mikau. And what I meant was that it's not like commutative or associative laws. If you don't know what L'H rule is, then you won't use it by accident, because it's not a common sense rule.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Right right. But there are still a few theorems and laws I've forgotten the names of but still use. But yeah I know what you mean.

A logarithm is just a misspelled algorithm.

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