- #1

- 421

- 1

Lim

_{X --> inf.}(1 - X + X

^{2})

^{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-->Y}g(x)

Lim

_{X-->Y}f(x).g(x) = Lim

_{X-->Y}f(x) . Lim

_{X-->Y}g(x)

This gives me

Lim

_{X --> inf.}X(1/X

^{2}- 1/X + 1)

^{1/2}- Lim

_{X --> 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.}