Can you spell the name of this function?

Calculus Level 5

Let ω ( x ) = n = 0 1 2 n sin ( 2 n x ) \omega(x) = \displaystyle \sum_{n=0}^{\infty} \dfrac{1}{2^{n}} \sin\left(2^{n}x \right) .

Evaluate lim h 0 ω ( x + h ) ω ( x ) h \displaystyle \lim_{h\rightarrow 0} \dfrac{\omega(x+h)-\omega(x)}{h} at x = 201 6 x=2016^{\circ} .

If the answer cannot be inputted into the solution box, select the correct option below and input the number of the option that best fits the correct answer:

  • Limit DNE, by approaching + + \infty on both sides of 2016: Input 1234

  • Limit DNE, by approaching - \infty on both sides of 2016: Input 4321

  • Limit DNE, by not approaching the same value from left and from right of 2016, but not approaching ± \pm \infty on either side: Input 1324

  • Limit DNE, by approaching -\infty from the left of 2016, and + + \infty from the right of 2016: Input 1243

  • Limit DNE, by approaching + +\infty from the left of 2016, and - \infty from the right of 2016: Input 3241


The answer is 1324.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Hobart Pao
Jan 8, 2016

This is a non rigorous solution. The function given is an example of Weierstrass Function, which is nowhere differentiable though it is everywhere continuous. If you graph the function, you'll notice that if you keep zooming in anywhere, there will be an infinite amount of bumps, but you're not going to have cusps. Thus, the answer is that the limit does not exist (because the derivative is a limit) and it doesn't matter where, because the limit does not exist anywhere, and it is by not having the same slope, approaching from either side of 2016, but not approaching positive or negative infinity.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...