Limits of trig functions

1. Oct 21, 2009

physicsman2

1. The problem statement, all variables and given/known data
lim as x approaches 2
(cos(pi/x))/(x-2)

lim as x approaches pi/4
(tan(x)-1)/(x-(pi/4))

2. Relevant equations

equations above

3. The attempt at a solution
for the first limit, i tried substituting t = (pi/2)-(pi/x) but i got stuck

i have no idea how to do the second limit

2. Oct 21, 2009

Staff: Mentor

Are you allowed to use L'Hopital's Rule?

3. Oct 21, 2009

physicsman2

no, i can only evaluate the limit as it is,

i was thinking of using t = pi/2 - pi/x and substituting the pi/x so i have an identity, but i don't think that will work

i have no idea on the second one, only changing tanx to sinx/cosx

4. Oct 21, 2009

Staff: Mentor

Well, phooey.
I don't have any ideas for the first one, but I believe the limit is pi/4 (using Excel to compute a few values for t near 2).

5. Oct 21, 2009

physicsman2

i was thinking of evaluating the limit of t at 2, so the answer would be zero, then substitute t in for pi/x and take the limit as t approaches 0

does this help?

6. Oct 21, 2009

Staff: Mentor

I don't understand what you're saying. Can you elaborate in the context of your problem?

7. Oct 21, 2009

Bohrok

For the first one, let u=$\pi$/x. Then use the identity cos u = sin($\pi$/2 - u) and rewrite the denominator so you have $\pi$/2 - u in it. Use another substitution with $\pi$/2 - u and you should get something like sinx/x. Don't forget to change the limits after a substitution.

8. Oct 21, 2009

physicsman2

well after saying t = pi/2 - pi/x, i would then evaluate the limit of t at x=2 and get 0, since pi/2 - pi/2 = 0

then, i would substitute the pi/x in the cos and put pi/2 - t in the cos function to get

(cos(pi/2 - t))/(x-2), then i would take the limit as t approaches 0 since the lim as x approaches 2 of t = 0. one problem is the x in the denominator, so then i would have to solve for x in t = pi/2 - pi/x

seeing that cos(pi/2 - t) is an identity, i would substitute that for the identity itself and solve but i can't get an answer

i also thought that a common identity is cos(pi/2 - theta) = sin theta, so i also tried to apply that to cos(pi/2 - t) to get sin(t), but i also don't get an answer

9. Oct 21, 2009

Staff: Mentor

This is what I was having trouble with. You don't take the limit of t at x=2. You take the limit of t, as t approaches something.

10. Oct 21, 2009

physicsman2

yeah i got my wording mixed up

bohrok, could you elaborate on your method a little more

thanks

11. Oct 21, 2009

Bohrok

u=$\pi$/x
limx→2 $\pi$/x = $\pi$/2, so use lim as u→$\pi$/2 in the new limit after the substitution.
$$\lim_{u\rightarrow \pi/2}\frac{\cos u}{\frac{\pi}{u} - 2} = \lim_{u\rightarrow \pi/2}\frac{\sin(\frac{\pi}{2} - u)}{\frac{\pi}{u} - 2} = \lim_{u\rightarrow \pi/2}\frac{\sin(\frac{\pi}{2} - u)} {\frac{2}{u}(\frac{\pi}{2} - u)}$$

Then use t = $\pi$/2 - u so you can finally use the limit sin(t)/t

12. Oct 21, 2009

Bohrok

The second one is involved, at least the way I did it.

First I wrote tanx as sinx/cosx and wrote the numerator as one fraction. The part I worked with that's making it indeterminate was (cosx-sinx)/(x-$\pi$/4)
Multiply by the conjugate and use the identity with cos2x, let t=2x, then use the identity cosθ = sin($\pi$/2 - θ). Use another substitution so you get a limit with sinx/x.

13. Oct 22, 2009

vertciel

Hi Bohrok,

I'm just evaluating this limit for fun. Could you please reveal the last substitution which you made to get $$\frac{\sin x}{x}$$?

Thanks.

14. Oct 22, 2009

Tedjn

Once you get to the equation with $\cos(2x)$, you can multiply top and bottom by 2 to get the desired form. There may be another substitution that would do the same thing.

15. Oct 22, 2009

Bohrok

$$\lim_{x\rightarrow \pi/4} \frac{\tan x - 1}{x - \frac{\pi}{4}} = \lim_{x\rightarrow \pi/4} \frac{\frac{\sin x}{\cos x} - \frac{\cos x}{\cos x}}{x - \frac{\pi}{4}} = \lim_{x\rightarrow \pi/4} \frac{\cos x - \sin x}{x - \frac{\pi}{4}} \times \frac{-1}{\cos x}$$

$$\frac{\cos x - \sin x}{x - \frac{\pi}{4}} \times \frac{\cos x + \sin x}{\cos x + \sin x} = \frac{\cos^2x - \sin^2x}{(x - \frac{\pi}{4})(\cos x + \sin x)} = \frac{\cos2x}{x - \frac{\pi}{4}} \times \frac{1}{\cos x + \sin x}$$

Let u = 2x and x = u/2. As x→$\pi$/4, u→$\pi$/2

$$\lim_{x\rightarrow \pi/4} \frac{\cos2x}{x - \frac{\pi}{4}} = \lim_{u\rightarrow \pi/2} \frac{\cos u}{\frac{u}{2} - \frac{\pi}{4}} = \lim_{u\rightarrow \pi/2} \frac{\sin(\frac{\pi}{2} - u)}{(-\frac{1}{2})(\frac{\pi}{2} - u)}$$

Let t = $\pi$/2 - u. As u→$\pi$/2, t→0

$$\lim_{u\rightarrow \pi/2} \frac{\sin(\frac{\pi}{2} - u)}{\frac{\pi}{2} - u} = \lim_{x\rightarrow 0} \frac{\sin t}{t}$$

And that takes care of the 0/0 part of the limit. Put everything together and you can find the limit.