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

[Helly123]

Homework Statement

$\lim_{h \to 0} \frac{f(x - 2h) - f(x + h)}{g(x + 3h) - g(x-h)}$

While f(x) = cos x
g(x) = sin x

Homework Equations

The Attempt at a Solution

Using L Hopital i couldn't make it more simple.
I tried to divide it by cos and sin
Can you give me clue?

 

[BvU]

Ever hear of Taylor series ?

 

[Ray Vickson]

Homework Statement

$\lim_{h \to 0} \frac{f(x - 2h) - f(x + h)}{g(x + 3h) - g(x-h)}$

While f(x) = cos x
g(x) = sin x

Homework Equations

The Attempt at a Solution

Using L Hopital i couldn't make it more simple.
I tried to divide it by cos and sin
Can you give me clue?

Show your work; we need to see some actual efforts (those are the rules in this forum).

 

[Helly123]

Ever hear of Taylor series ?

I never heard of that. but, I will look for it

 

[Helly123]

Show your work; we need to see some actual efforts (those are the rules in this forum).

ok. I tried this :
1)
$\lim_{h \to 0} \frac{\cos(x - 2h) - \cos(x + h)}{\sin(x + 3h) - \sin (x - h)}$
$cos(a \pm b) = \cos a cos b \mp sin a sin b$
$\lim_{h \to 0} \frac{\cos(x)\cos(2h) + \sin(x)\sin(2h) - \cos(x)\cos(h) + \sin(x)\sin(h)} {\sin(x + 3h) - \sin (x - h)}$

$\lim_{h \to 0} \frac{\cos(x)(\cos(2h) - \cos(h) ) + \sin(x)( \sin(2h) + \sin(h) )} {\sin(x + 3h) - \sin (x - h)}$​

2)
$\lim_{h \to 0} \frac{\cos(x - 2h)- \cos(x + h)}{\sin(x + 3h) - \sin (x - h)} * \frac{\cos(x - 2h) + \cos(x + h)}{\cos(x - 2h) + \cos(x + h)} * \frac {\sin(x + 3h) + \sin (x - h)}{\sin(x + 3h) + \sin (x - h)}$
$\lim_{h \to 0} \frac{ (\cos(x - 2h)^2 - \cos(x + h)^2 ) (\sin(x + 3h) + \sin (x - h) )}{(\sin(x + 3h)^2 - \sin (x - h)^2) (\cos(x - 2h) + \cos(x + h) ) }$

3) $\lim_{h \to 0} \frac{\cos(x - 2h) - \cos(x + h)}{\sin(x + 3h) - \sin (x - h)}$
L Hospital rule
$\lim_{h \to 0} \frac{-\sin(x - 2h) + \sin(x + h)}{\cos(x + 3h) - \cos (x - h)}$
should I derive it with respect to x or h?

 

[BvU]

1) is the good path. What does the numerator give you in first order of $h$ ?
Then do same for denominator and bingo.

3) wrt x but you end up doing the same work as in 1).

 

[Dick]

3) wrt x but you end up doing the same work as in 1).

No, to apply l'Hopital to this problem you differentiate with respect to the variable in the limit. That's $h$.

 

[Ray Vickson]

ok. I tried this :
1)
$\lim_{h \to 0} \frac{\cos(x - 2h) - \cos(x + h)}{\sin(x + 3h) - \sin (x - h)}$
$cos(a \pm b) = \cos a cos b \mp sin a sin b$
$\lim_{h \to 0} \frac{\cos(x)\cos(2h) + \sin(x)\sin(2h) - \cos(x)\cos(h) + \sin(x)\sin(h)} {\sin(x + 3h) - \sin (x - h)}$

$\lim_{h \to 0} \frac{\cos(x)(\cos(2h) - \cos(h) ) + \sin(x)( \sin(2h) + \sin(h) )} {\sin(x + 3h) - \sin (x - h)}$​

2)
$\lim_{h \to 0} \frac{\cos(x - 2h)- \cos(x + h)}{\sin(x + 3h) - \sin (x - h)} * \frac{\cos(x - 2h) + \cos(x + h)}{\cos(x - 2h) + \cos(x + h)} * \frac {\sin(x + 3h) + \sin (x - h)}{\sin(x + 3h) + \sin (x - h)}$
$\lim_{h \to 0} \frac{ (\cos(x - 2h)^2 - \cos(x + h)^2 ) (\sin(x + 3h) + \sin (x - h) )}{(\sin(x + 3h)^2 - \sin (x - h)^2) (\cos(x - 2h) + \cos(x + h) ) }$

3) $\lim_{h \to 0} \frac{\cos(x - 2h) - \cos(x + h)}{\sin(x + 3h) - \sin (x - h)}$
L Hospital rule
$\lim_{h \to 0} \frac{-\sin(x - 2h) + \sin(x + h)}{\cos(x + 3h) - \cos (x - h)}$
should I derive it with respect to x or h?

To finish (1), you need to replace $\cos(2h), \cos(h), \sin(2h)$ and $\sin(h)$ by their lowest-order approximations in small $h$. If you look at a unit circle, centered at the origin, with a point on it at $(\cos(w), \sin(w))$ for small angle $w$ you should be able to see "geometrically" what are $\cos(w)$ and $\sin(w)$ for small $w$ up to terms linear in $w$. In other words, if we set $\cos(w) = a + bw$ and $\sin(w) = c + dw$ for small $w$, you can just look at the geometry to figure out the values of $a, b, c, d$. Then, replace the sin and cosine of the small angles by their linear small-angle forms.

Finally, as BVU suggests, do the same thing in the denominator.

 

[BvU]

No, to apply l'Hopital to this problem you differentiate with respect to the variable in the limit. That's $h$.

My bad ! Thank you. And then I should also retract my second claim there: it gives the right answer and all you have to do is set $h$ to 0.

Goes to show you're never too old to learn (don't recall ever learning about L'Hopital and have never used it either).

 

[SammyS]

Goes to show you're never too old to learn

(don't recall ever learning about L'Hopital and have never used it either). ​

​

That seems reasonable. It would have been difficult to use it if you had never learned it.

 

[BvU]

That seems reasonable. It would have been difficult to use it if you had never learned it.

Perhaps unknowingly, like when you turn to Taylor series to find a limit ?

 

[Helly123]

I ended up with :
$\frac{\cos(x)( \cos (2h) - \cos(h) ) + \sin (x) (\sin (2h) + \sin(h)) } {\sin(x) (\cos(3h) - \cos(h)) + \cos (x)(\sin(3h) + \sin(h))}$
But, how to get 0/0 result disappear?

 

[SammyS]

I ended up with :
$\frac{\cos(x)( \cos (2h) - \cos(h) ) + \sin (x) (\sin (2h) + \sin(h)) } {\sin(x) (\cos(3h) - \cos(h)) + \cos (x)(\sin(3h) + \sin(h))}$

You have done incorrect differentiation.

 

[Helly123]

To finish (1), you need to replace $\cos(2h), \cos(h), \sin(2h)$ and $\sin(h)$ by their lowest-order approximations in small $h$. If you look at a unit circle, centered at the origin, with a point on it at $(\cos(w), \sin(w))$ for small angle $w$ you should be able to see "geometrically" what are $\cos(w)$ and $\sin(w)$ for small $w$ up to terms linear in $w$. In other words, if we set $\cos(w) = a + bw$ and $\sin(w) = c + dw$ for small $w$, you can just look at the geometry to figure out the values of $a, b, c, d$. Then, replace the sin and cosine of the small angles by their linear small-angle forms.

Finally, as BVU suggests, do the same thing in the denominator.

Can we just use.. more high-school way of solving problem. My brain cannot connect to the professor method.
(Pardon my English)

 

[Helly123]

You have done incorrect differentiation.

Which is wrong Sir?

 

[SammyS]

Which is wrong Sir?

What is the derivative of $\ \cos(2h) \$ for example?

 

[Helly123]

What is the derivative of $\ \cos(2h) \$ for example?

How if i solve it with method 1 in #5 post? By L Hospital the answer is $\frac{3}{4} \tan x$

 

[SammyS]

How if i solve it with method 1 in #5 post? By L Hospital the answer is $\frac{3}{4} \tan x$

Do it by steps.
Your derivatives are not correct for numerator and denominator.

From Post #5

$\displaystyle \lim_{h \to 0} \frac{\cos(x)\cos(2h) + \sin(x)\sin(2h) - \cos(x)\cos(h) + \sin(x)\sin(h)} {\sin(x + 3h) - \sin (x - h)}$​

Taking the derivative of the numerator with respect to h is not

$\cos(x)( \cos (2h) - \cos(h) ) + \sin (x) (\sin (2h) + \sin(h))$​

You have a similar problem with the denominator.

Please give more detail on what you are doing, rather than having us guess, (although I did guess correctly).

You are NOT applying the chain rule at all! You need to apply it and apply it correctly.

 

[Helly123]

Do it by steps.
Your derivatives are not correct for numerator and denominator.

From Post #5

$\displaystyle \lim_{h \to 0} \frac{\cos(x)\cos(2h) + \sin(x)\sin(2h) - \cos(x)\cos(h) + \sin(x)\sin(h)} {\sin(x + 3h) - \sin (x - h)}$​

Taking the derivative of the numerator with respect to h is not

$\cos(x)( \cos (2h) - \cos(h) ) + \sin (x) (\sin (2h) + \sin(h))$​

You have a similar problem with the denominator.

Please give more detail on what you are doing, rather than having us guess, (although I did guess correctly).

You are NOT applying the chain rule at all! You need to apply it and apply it correctly.

I get your point..
By L Hospital :
And applying chain rule

$\lim_{h \to 0} \frac{\cos(x-2h) - \cos (x + h)}{\sin (x + 3h) - \sin(x - h)}$

$\lim_{h \to 0} \frac{-2 . -\sin(x-2h) + \sin (x + h)}{3. \cos (x + 3h) + \cos(x - h)}$

$\frac{2\sin(x) + sin(x)}{3\cos (x) + \cos(x)}$

$\frac{3}{4} \tan x$

 

[Helly123]

I want to solve it without derivation..

 

[SammyS]

I want to solve it without derivation..

Differentiation or Differentiating .

I want to solve it without derivation..

What is stopping you ?

 

[Helly123]

Differentiation or Differentiating .What is stopping you ?

I did with differentiation right? I want to know different way to solve it, without L Hospital..

 

[Ray Vickson]

I did with differentiation right? I want to know different way to solve it, without L Hospital..

So, follow the method suggested in post #8!

 

[Helly123]

So, follow the method suggested in post #8!

But i seriously don't get it..

 

[Helly123]

So, follow the method suggested in post #8!

Can you explain more?

 

[BvU]

Do you know about $$\lim_{h\downarrow 0} {\sin h\over h} \ \ ? $$ Without Taylor (too close to differentiation for your liking) you are going to need it.

 

[Helly123]

Do you know about $$\lim_{h\downarrow 0} {\sin h\over h} \ \ ? $$ Without Taylor (too close to differentiation for your liking) you are going to need it.

Yea.. it is 1

 

[BvU]

Good.
You'll also need $\displaystyle { \lim_{h\downarrow 0} {\cos h\over h^2}} \ \$ I'm afraid.

[edit] glad you like it, but it doesn't exist and I should correct:
$$\lim_{h\downarrow 0} {\cos h - 1 \over h^2}$$is what I should have written ...​

Work out the denominator in #5 1) in the same way as the numerator.

Cheat: ${3\over 4} \tan x\$ is correct, so you know that you want to work toward a sine in the numerator and a cosine in the denominator. The 3 and the 4 are a lot of work this way, though...(so I'm glad I had Taylor in my toolbox and now I have this hospital guy too. I'm happy! )

 

[Ray Vickson]

Can you explain more?

Yes. Just do what I suggested: look at the "geometry", to find the constants $a,b,c,d$ in the linear approximations $\sin(w) = a + bw$ and $\cos(w) = c + dw$ for small angles $w$.

 

[epenguin]

I want to solve it without derivation..

# 5 1) and #12 are on track and you and using the "small angle formulae", #27, 28 you can complete.

 

[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.