- #1
WiFO215
- 420
- 1
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.
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.