How Does L'Hopital's Rule Solve This Limit Problem?

[cummings15]

Homework Statement

lim (sin4x) divided by (x^2+8x)
x approaches 0

Homework Equations

L'Hopital's rule

The Attempt at a Solution

u = sin4x
du = cos4x

y = x^2+8x
dy = 2x+c

 

[LCKurtz]

Homework Statement

lim (sin4x) divided by (x^2+8x)
x approaches 0

Homework Equations

L'Hopital's rule

The Attempt at a Solution

u = sin4x
du = cos4x

The derivative would be called du/dx, not just du. And cos(4x) isn't correct. Remember the chain rule.

y = x^2+8x
dy = 2x+c

Again you mean dy/dx and the derivative isn't correct. And where did the + c come from? You aren't integrating.

Finally, once you correct the derivatives you need to write something like this:

$$lim_{x\rightarrow 0}\frac {\sin(4x)}{x^2+8x} = \ ?$$

and write on the right side what you get.

 

[Mark44]

Homework Statement

lim (sin4x) divided by (x^2+8x)
x approaches 0

Homework Equations

L'Hopital's rule

The Attempt at a Solution

u = sin4x
du = cos4x

This is not right. You forgot to use the chain rule, plus it's missing the dx on the right side.

y = x^2+8x
dy = 2x+c

This is also incorrect. The derivative of 8x is 8, not c. You're also missing the dx.

In any case, you need to be working with derivatives, not differentials.

What do you get when you actually apply L'Hopital's Rule? This rule says that under certain conditions,
$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$
If the latter limit exists, the first limit is equal to it.