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.