# Infinitesimal calculus problem can't seem to solve

1. Nov 16, 2008

### watr

Hi all,

I have been working on this problem for the past 2 days and can't seem to get it. I have gotten parts of the answer but not the answer. Could someone explain this to me?

The answer is sqrt(2) - 1 or 1 / ( 1 + sqrt(2) )

Problem: H is infinity
sqrt(H+1) / ( sqrt(2H) + sqrt(H-1) )

I am completely stumped. The best thing I have done in getting close to the answer was to take out the sqrt(2) out of the sqrt(2H), and then to multiply top and bottom by
sqrt(H) + sqrt(H-1). My main problem was that I couldn't simplify the problem down to the answer. Maybe I only need to simplify it to some point and then use the finite math theorems...any ideas would be greatly appreciated.

Note, if you could point out some useful hints, that would be preferable to just solving it for me.

Cheers

2. Nov 16, 2008

### HallsofIvy

This makes no sense at all! "infinity" is not a real number so you cannot take "H= infinity".
Is it at all possible that the problem REALLY is
$$\lim_{H\rightarrow \infty} \frac{\sqrt{H+1}}{\sqrt{2H}+ \sqrt{H-1}}$$?

If so, since "infinity" is hard work with- as I said, it is not a real number so you can't just put in "H= infinity"- I would divide both numerator and denominator by $\sqrt{H}$, taking that inside each squareroot as just H. You should get fractions with H in the denominator and that goes to 0.

3. Nov 16, 2008

### natives

I definitely agree with the notion that the question was posted in faulty grammar "H is infinity or H=infinity" confuses the whole thing..simple grammatical but great mathematical illusional consequences...in answering to your question,yes indeed it is a question of limits and to get that answer then it should be written as lim H-->infinity then the function f(H) as in the question...also factor out sqroot(H) from both denominator and numerator,during this operation DO NOT remember that H goes to infinity,forget that..take it as a normal variable number then do the math!Then u'd have sqroot(1+1/H)/[sqroot(2)+sqroot(1-1/H)]...then NOW is when you remember that H actually goes to infinity,a very large number mind you so the fractions 1/H will tend to go down to zero,then you have your 1st answer = 1/(sqroot(2)+1)..to get the other alternative answer just multiply the above with (sqroot(2)-1)/(sqroot(2)-1) then you'd have the 2nd answer = sqroot (2)-1....where sqroot(x)=square root of x

4. Nov 16, 2008

### watr

Thanks for the responses, that makes a lot of sense.

As for the H = infinity, there is no = sign with the squigily lines as ~ so I am not sure how to show that H is approaching infinity without using a limit here.

Thanks for the responses.

5. Nov 17, 2008

### HallsofIvy

Then just say "as H goes to 0"!