Monday, May 15, 2006

I Hope That's All. Topic: Calculus. Level: AIME.

Problem: (Problem-Solving Through Problems - 6.7.6) Calculate

$ \displaystyle \lim_{x \to \infty} \left(x \int_0^x e^{t^2-x^2} dt \right) $.

Solution: First, we want to simplify it. Note that $ x $ is not the variable of integration, which means it can be regarded as a constant. In particular, we have

$ \displaystyle x \int_0^x e^{t^2-x^2} dt = xe^{-x^2} \int_0^x e^{t^2} dt $.

We notice that $ e^{t^2} $ is not fun to integrate, so we think L'Hopital's Rule. Rewrite the limit as

$ \lim_{x \to \infty} \left(\frac{\int_0^x e^{t^2} dt}{\frac{e^{x^2}}{x}} \right) $.

This is a $ \frac{\infty}{\infty} $ indeterminate form, so we can apply L'Hopital's Rule. Our limit becomes (after differentiating the top and the bottom)

$ \displaystyle \lim_{x \to \infty} \left(\frac{e^{x^2}}{\frac{2x^2e^{x^2}-e^{x^2}}{x^2}}\right) = \lim_{x \to \infty} \left(\frac{x^2}{2x^2-1}\right) = \frac{1}{2} $.



Practice Problem: (Problem-Solving Through Problems - 6.7.2) Suppose that $ f $ is a function with two continuous derivatives and $ f(0) = 0 $. Prove that the function $ g $ defined by

$ g(0) = f^{\prime}(0) $, $ g(x) = \frac{f(x)}{x} $ for $ x \neq 0 $

has a continuous derivative.


  1. I didn't get the first part but..
    if g(0) is equal to 0, then f ^1 would also be 0, cuz they're equal. Then that means f(x) is also equal to 0, cuz anything ^1 is equal to the same thing. that leaves g(x)=0/x. 0 divided by anything is 0, therefore g(x)=0. And since g(0)=f(0), then f(0) is also equal to 0.
    .. well, i tried.....

  2. That's actually f'(0) as in f-prime, not f^1. It's a derivative, aka calculus.

  3. yay for trying ^^ that's more than i do....

  4. although i get it... i could never do that but i get it....

  5. lim_{x \to 0} f(x) / x = f'(0)

    Thus g(x) is continuous. g'(x) has a removable singularity at 0, and since f is twice differentiable it follows g'(x) is continuous everywhere.

  6. Well, whoops, that proof's not really complete.

    g'(x) = - f(x) / x^2 + f'(x) / x = ( x f'(x) - f(x) ) / x^2

    \lim_{x \to 0} g'(x) = (x f''(x) + f'(x) - f'(x) ) / 2x = f''(x) / 2

    So the limit is obviously defined on both sides and hence g'(x) is continuous.

  7. AIME level? I thought calculus wasn't on AIME ... And if it was ... would this be question 1 or something?

  8. Calculus isn't on the AIME... but I don't have another system for difficulty. It'd probably be an easy AIME question, yeah.

  9. i don't get the title... :?