L'Hopital's Rule: Solving Homework Statement

[Nope]

Homework Statement

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

Homework Equations

The Attempt at a Solution

answer is 243/25
The derivative of sin 5x-5x is always 0,
dunno know how to do it...

 

[Inirit]

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.

 

[Nope]

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

 

[Deadstar]

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?

 

[Inirit]

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.

 

[Deadstar]

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...

 

[Inirit]

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.

 

[Nope]

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

 

[Inirit]

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.

 

[Deadstar]

Differentiate again and it will work.

 

[Nope]

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

 

[Nope]

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]

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.

 

[Deadstar]

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