Something's wrong

  • Thread starter WiFO215
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  • #1
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I was given this review problem in limits.

LimX --> inf. (1 - X + X2)1/2 - aX - b = 0.

I was asked to find a and b such that the above equation is satisfied, which I did as follows:

I removed an X from the entire thing and expanded the term in the brackets using binomial theorem. I get a = 1 and b = -1/2 which is the answer given in the text.

But here's what is bothering me. I did the sum another way and got a different answer and can't put my finger on the mistake.
I remove an X from the root term. And I split the limit across the functions since

Lim X-->Y f(x) + g(x) = Lim X-->Y f(x) + Lim X-->Yg(x)

Lim X-->Yf(x).g(x) = Lim X-->Yf(x) . Lim X-->Y g(x)

This gives me

LimX --> inf. X(1/X2 - 1/X + 1)1/2 - LimX --> inf. (aX + b) = 0

= [LimX --> inf. X] [LimX --> inf.(1/X2 - 1/X + 1)1/2] - LimX --> inf. (aX + b) = 0

Now the term inside the root sign goes to 1. We are left with
LimX --> inf. X - aX - b = 0

This way, the answer is a = 1 and b = 0.

Why am I getting a different answer? I used all the limit rules correctly as far as I can see.
 

Answers and Replies

  • #2
1,101
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I think you are haphazardly applying the formulas for the algebraic operations on limits. Remember, you can split limits across functions provided that all limits in question actually exist. For example, I don't think you can split the square root term as you did since X goes off to infinity.
 
  • #3
258
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The second way you did it makes it so the 1 - x term disappears. Intuitively, the 1 disappearing is fine since when it comes to infinity, it's meaningless, but the x can't disappear like that.
 
  • #4
726
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First, [tex] \lim_{x \rightarrow \infty} x\sqrt[]{(1/x^2 - 1/x + 1)} - ax - b = \lim_{x \rightarrow \infty} x\sqrt[]{(1/x^2 - 1/x + 1)} - (ax + b) [/tex].

If this limit is to go to 0, then obviously a must be positive (for if a were negative the entire expression we're trying to find the limit of would increase beyond all positive bounds). But if a is positive, then the expression above is of the indeterminate form "infinity - infinity", and so you can't say that the square root expression tends to 1 and the x's will then cancel out.
 
  • #5
421
1
Remember, you can split limits across functions provided that all limits in question actually exist. For example, I don't think you can split the square root term as you did since X goes off to infinity.

D'oh!! How could I forget that??
 

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