Find the limit as h --> 0 for this trigonometery equation

[mattt]

You only need to know that \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = f'(x), if you know that, you can solve it (you don't need to know about Taylor Series, L'Hopital,...), just "transform" what you have, to express it in terms of the limit definition of diferentiation.

 

[mattt]

For example, f(x-2h)-f(x+h)=f(x-2h)-f(x)+f(x)-[f(x+h)-f(x)+f(x)]=f(x-2h)-f(x)-[f(x+h)-f(x)]

=-2h*\frac{f(x-2h)-f(x)}{-2h}-\left(h*\frac{f(x+h)-f(x)}{h}\right)=

=-h*\left(2*\frac{f(x-2h)-f(x)}{-2h}+\frac{f(x+h)-f(x)}{h}\right)

You must do the same thing in the denominator, then simplify, and then the limit will be clear.

 

[SammyS]

For example, f(x-2h)-f(x+h)=f(x-2h)-f(x)+f(x)-[f(x+h)-f(x)+f(x)]=f(x-2h)-f(x)-[f(x+h)-f(x)]=

=-2h*\frac{f(x-2h)-f(x)}{-2h}-\left(h*\frac{f(x+h)-f(x)}{h}\right)=

=-h*\left(2*\frac{f(x-2h)-f(x)}{-2h}+\frac{f(x+h)-f(x)}{h}\right)

You must do the same thing in the denominator, then simplify, and then the limit will be clear.

Beside the fact that this is the pre-calculus section, it looks like you propose to differentiate the numerator and denominator. That implies using L'Hospital's rule.

 

[mattt]

Beside the fact that this is the pre-calculus section, it looks like you propose to differentiate the numerator and denominator. That implies using L'Hospital's rule.

First, I am not differentiating anything.
Second, I am not using L'Hopital.

Just read what I wrote (simple arithmetic equalities). Do you understand that each "=" sign is correct?

 

[SammyS]

First, I am not differentiating anything.
Second, I am not using L'Hopital.

Just read what I wrote (simple arithmetic equalities). Do you understand that each "=" sign is correct?

If I understand your idea, you are correct to say that you are not applying L'Hopital .

If I'm right about where this is headed, it still requires recognizing and knowing derivatives of sine and cosine.

... but that (your method) is a nice way to look at this.

 

[mattt]

If I understand your idea, you are correct to say that you are not applying L'Hopital .

If I'm right about where this is headed, it still requires recognizing and knowing derivatives of sine and cosine.

If you use my way of solving it, you will not need to even know that f(x)=cos(x) and g(x)=sin(x), the result does not depend on that (in fact, the result is \frac{-3 f'(x)}{4 g'(x)} regardless of who is f(x) and g(x).

To know why it is so, you only need to know that, in general, \frac{f(x+A)-f(x)}{A} goes to f'(x) when A goes to zero (whatever expression A may be).

 

[epenguin]

The OP said he wants to get the conclusion without using calculus, but is using the specific f(x) and g(x) of the original problem in #1, and is quite close

 

[SammyS]

If you use my way of solving it, you will not need to even know that f(x)=cos(x) and g(x)=sin(x), the result does not depend on that (in fact, the result is \frac{-3 f'(x)}{4 g'(x)} regardless of who is f(x) and g(x).

... and we know that $\displaystyle \ \frac{-3 f'(x)}{4 g'(x)} = \frac{3}{4} \tan(x) \$ because ____ ?

 

[mattt]

... and we know that $\displaystyle \ \frac{-3 f'(x)}{4 g'(x)} = -\frac{3}{4} \tan(x) \$ because ____ ?

What I wrote in #32 are simple arithmetic manipulations (in the numerator of the fraction) to get to -h*\left(2*\frac{f(x-2h)-f(x)}{-2h} + \frac{f(x+h)-f(x)}{h}\right)

If you do the same type of "clever" arithmetic manipulations in the denominator, you will get to: h*\left(3*\frac{g(x+3h)-g(x)}{3h} + \frac{g(x-h)-g(x)}{-h}\right)

In short:

\frac{f(x-2h)-f(x+h)}{g(x+3h)-g(x-h)} = \frac{-h*\left(2*\frac{f(x-2h)-f(x)}{-2h} + \frac{f(x+h)-f(x)}{h}\right)}{h*\left(3*\frac{g(x+3h)-g(x)}{3h} + \frac{g(x-h)-g(x)}{-h}\right)}=

=\frac{-\left(2*\frac{f(x-2h)-f(x)}{-2h} + \frac{f(x+h)-f(x)}{h}\right)}{\left(3*\frac{g(x+3h)-g(x)}{3h} + \frac{g(x-h)-g(x)}{-h}\right)}=

And the limit of that last mathematical expression, when h goes to zero, is:

=\frac{-3*f'(x)}{4*g'(x)}.

 

[SammyS]

=\frac{-3f'(x)}{4g'(x)}.

I have no argument with getting that far.

It's just not the final result.

 

[mattt]

I have no argument with getting that far.

It's just not the final result.

Sorry, I don't understand what you mean.

I just gave a mathematical proof of \lim_{h\to 0}\frac{f(x-2h)-f(x+h)}{g(x+3h)-g(x-h)}=\frac{-3*f'(x)}{4*g'(x)} for any "f" and "g" with the sole condition that both are differentiable at the point "x".

If you want to particularize that general result, for the concrete case f(x) = cos(x) and g(x) = sin(x), then:

\lim_{h\to 0}\frac{f(x-2h)-f(x+h)}{g(x+3h)-g(x-h)}=\frac{-3*f'(x)}{4*g'(x)}=\frac{3*sin(x)}{4*cos(x)}=\frac{3}{4}*tan(x)