# Limits, L'Hospital's Rule & Trigonometry

1. Oct 26, 2008

### CougarXLS

SOLVED, thanks to Dick and viciousp!

1. The problem statement, all variables and given/known data
The question asks: Find the limit. Use L'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If L'Hospital's Rule doesn't apply, explain why.

Limit as x approaches pi/2 from the right (+) of given equation:
lim [x -> (pi/2)+] of [(cos x) / (1 - sin x)]

2. Relevant equations
Possibly trig identity: tan x = sin x / cos x
L'Hosptial's Rule: lim(x->a) of f(x)/g(x) = lim(x->a) of fprime(x) / gprime(x)

3. The attempt at a solution
lim [x -> (pi/2)+] of [(cos x) / (1 - sin x)]
= lim [x -> (pi/2)+] of [(-sin x) / (-cos x)] ---> using L'Hosptial's Rule
= lim [x -> (pi/2)+] of [(tan x)] ---> using trig identity tan x = sin x / cos x

--> tan (pi/2) is undefined. Therefore my solution would be infinity.
I don't understand how the answer is negative infinity. LOL, my answer disagreeing with the book again... lol... I know it's probably me.

Could someone please explain to me how it's negative infinity and not infinity?

Last edited: Oct 26, 2008
2. Oct 26, 2008

### Dick

Think about it. If x is a little bigger than pi/2 then -sin(x) is approximately -1, and cos(x) is a small negative number, so -cos(x) is a small positive number. -1/(small positive) -> -infinity. Seeing the tan(x) is undefined tells you it's one of the infinities, but doesn't tell you which.

3. Oct 26, 2008

### CougarXLS

But it's not -sin x.... the negatives canceled out...

lim [x->pi/2]+ of -sinx / -cosx leaves a positive (the -1 cancels out)

I don't understand. Even if (pi/2) is slightly bigger (as approaching from right), then sin x still goes to 1, not -1.

I don't mean to be difficult; I wish to understand.
Thanks.

4. Oct 26, 2008

### Dick

You are just moving the -1 around. Ok, sin(x) goes to 1. cos(x) is a small negative. 1/(small negative) -> -infinity.

5. Oct 26, 2008

### CougarXLS

Forgive me, but if the negatives canceled out, how do I know to 'move the -1' around?

I realize that tan x is undefined and that it tells you it's one of the infinities, but doesn't tell you which.... from that I concluded that because the -1 canceled out, the resulting limit would be positive.

Wow, I am confused.... :)

6. Oct 26, 2008

### Dick

Limit x->pi/2+ of tan(x) is -infinity. Because sin(x)->1 (as you said) and cos(x) is a small negative number. Just because cos(x) doesn't have a '-' in front of it doesn't mean it can't be negative. Look at a graph of cos(x).

7. Oct 26, 2008

### CougarXLS

Okay, what you say makes sense, but how do I know which it is?

When I look at the graph of f(x)=cos x, I can understand how it can yield a negative value. What I don't understand is how, from the equation I am finding the limit of, how can I determine that a -1 is still hanging around?
Is there some trig trick/identity/rule I am missing?

8. Oct 26, 2008

### viciousp

cos(pi/2)=0

between 0 and pi/2 cos is positive
between pi/2 and pi cos is negative

cos(pi)=-1

Since we are coming from the right the value of cosine is always negative but increasing (getting closer to zero)

9. Oct 26, 2008

### Dick

The original limit has the form 0/0. That means you can apply l'Hopital to the limit. You got tan(x). The limit of tan(x) as x->pi/2+ is minus infinity. l'Hopital's rule then tells you that is also the value of the original limit.

10. Oct 26, 2008

### CougarXLS

Yes, I am with you so far.

In typing this response, I seem to have neglected the observation that I am taking the limit from the right and that All Students Take Calculus.... aka tan (pi/2 + 0.01) < 0 (where 0.01 is some very small positive number)! Which of course is negative.
Sometimes I think I am in over my head. lol

Lol, that's the one part of this section I understand.

Thanks Dick and viciousp (for reminding me of C.A.S.T., it was your post that jogged that old omen). I get it know.

What can I say... lol, it's been four years out of high school and into first year university.