# L'Hopital's Rule

## Homework Statement

lim x->0
5x(cos 9x-1)/sin 5x-5x

## The Attempt at a Solution

The derivative of sin 5x-5x is always 0,
dunno know how to do it...

Is it sin(5x-5x) or sin(5x)-5x? The first one seems like a trivial way to express an equation and would result in 0 anyway, making the original equation undefined.

I'm not sure, it's an assignment from the internet,
it should be sin(5x-5x) , cause it ask you to use L'Hopital's Rule
sin(0)=0

Has to be

$$\lim_{x \to 0} \frac{5x(\cos(9x) - 1)}{\sin(5x) - 5x}$$

OP the derivative of $$5x(\cos(9x) - 1)$$ is

$$5\cos(9x) - 5 - 45x \sin(9x)$$

and the derivative of $$\sin(5x) - 5x$$ is

$$5\cos(5x) - 5$$

Know how to proceed from here?

You'd use the rule if you got a 0 in the denominator after plugging in the limit of x, not if it were 0 to begin with. An equation with a 0 in the denominator would always have a 0 in the denominator no matter how many times you applied the rule.

Last edited:
You'd use the rule if you got a 0 in the denominator after plugging in the limit of x, not if it were 0 to begin with. An equation over 0 would always be over 0 no matter how many times you differentiate it.

Not quite...

An equation over zero is undefined.

x/0 is not equal to zero.

Edit: Actually I think you meant if you differentiate zero you get zero...

Edit: Actually I think you meant if you differentiate zero you get zero...

Yes, that's what I meant, but I see that I worded it incorrectly before. To explain a little better, though, I meant that if you applied the rule to an equation over 0 (f(x)/0), it would still be over 0 no matter how many times you differentiated the top and bottom.

$$\lim_{x \to 0} \frac{5x(\cos(9x) - 1)}{\sin(5x) - 5x}$$
I tried
After the second derivative, the denominator is still zero...
-25sin(5x)

What's to stop you from differentiating a third time? Without the extra factor, it looks like the denominator would turn into a function of cosine, which wouldn't give you zero when you plug in zero.

Differentiate again and it will work.

ok ty, I will do it later, I have class now

I differentiated like 5 times, I still got 0..
I notice that there is always a x*cos(9x) in the nominator...which make whole thing equal to 0..

HallsofIvy
Homework Helper
Yes, there is nothing at all wrong with getting "0" in the numerator. As long as the denominator is not also 0, that just tells you that the limit is 0.

I differentiated like 5 times, I still got 0..
I notice that there is always a x*cos(9x) in the nominator...which make whole thing equal to 0..

Then you've made a mistake somewhere, can get messy at times.

First derivatives.

$$\frac{5 \cos(9x) - 5 - 45x\sin(9x)}{5\cos(5x) - 5}$$

Second one

$$\frac{90 \sin(9x) + 405x\cos(9x)}{25\sin(5x)}$$

third one

$$\frac{1215 \cos(9x) + 3645x\sin(9x)}{125\cos(5x)}$$

Now sub in x = 0