Wednesday, May 17, 2006

Calculus Problem #1: Limit Challenge

OK, my illustrious, math-proficient readers, here is your first challenge!

Before you learn how to do derivaties in the normal way, you calculate them using limits. The general formula looks like this:If you try this for, for instance, f(x) = x^2, you'll see you can pretty easily get to f'(x) = 2x. (The general trick, if this all looks completely unfamiliar, is to somehow massage the formula until it's solvable with h = 0.)

The one I couldn't solve was the one where f(x) = sqrt(1 + 2x). The formula for the limit looks like this:
I'm pretty sure I'll have to multiply by some kind of square root to get the thing solved, but so far no luck. (As a reminder, I do know the actual answer - this is an odd-numbered question, after all - so what I'm looking for is how to actually calculate this limit.)

7 comments:

cartaufalous said...

Multiply top and bottom by 1 over the square root of 1+2x.

cartaufalous said...

Which lets me get rid of the 2, the 4x, and the 2h, but they're not the terms that matter. So I'm left with no fractions, but too many square roots still. Hmmmm . . .

cartaufalous said...

Three hours later . . .

Nothing to it! You just seperate out all the "h" terms until you get something that doesn't blow up. I think the answer is 1 over 2 times the square root of 1 + 2x. But after eleven pages of chickenscribble it's hard to say for sure. There's got to be an easier way.

Tam said...

Your answer is nearly right but not quite - I suspect a simple math error somewhere.

But now the question is, can you (or someone else) explain how to get there (hopefully without 11 pages of work!)

Anonymous said...

Multiply both denominator and numerator by [sqrt(1 + 2(x+h) ) + sqrt(1+2x) ]

King Al I said...

Multiply both denominator and numerator by [sqrt(1 + 2(x+h) ) + sqrt(1+2x) ]

cartaufalous said...

Well, that sure beats the brute-force method.