1. ## Finding the Limit

My problem is this:

I've tried this multiple times, and I've gotten (1/16) or -(1/16).

Using L'Hospital's Rule, the derivative of the top fuction was:

cos(pi*e^x^8), which I calculated to be -1 after I put 0 in.

The bottom derivative turned out to be:

-16/(16*x^2+8*x+1)

I plugged in 0, and I got -16.

My homework still tells me that I'm wrong.

2. Originally Posted by Kimmy2
My problem is this:

I've tried this multiple times, and I've gotten (1/16) or -(1/16).

Using L'Hospital's Rule, the derivative of the top fuction was:

cos(pi*e^x^8), which I calculated to be -1 after I put 0 in.

The bottom derivative turned out to be:

-16/(16*x^2+8*x+1)

I plugged in 0, and I got -16.

My homework still tells me that I'm wrong.

$\left(x\cos^5\left(\pi e^{x^8}\right)\right)'=\cos^5\left(\pi e^{x^8}\right)-40x^8e^{x^8}\pi \cos^4\left(\pi e^{x^8}\right)\sin\left(\pi e^{x^8}\right)$ and this thing goes to -1 when x goes to zero.

And the bottom's derivative is nothing close to what you say: $\left(\ln(1+4x)\right)'=\frac{4}{1+4x}\xrightarrow [x\to 0] {}4$ , so the limit is in fact $-\frac{1}{4}$

Tonio