# death by limits!!!

• September 4th 2006, 04:11 PM
sacwchiri
death by limits!!!
hello... i just started university and in my math class we are seing limits and since here it said limits well i decided to post it here....

well ive been having lots of trouble with trigonometry stuff and now they go and mix one problem with limits

Im supposed to find:

lim {cos^2(x/2)}/{cos(x)-sen(3x/2)}
x->pi

thanks for the help...
• September 4th 2006, 05:16 PM
ThePerfectHacker
What does "sen" mean? secant?
• September 4th 2006, 05:27 PM
sacwchiri
death by limits!!!
amm yeah... sen => sin ...... its in spanish and forgot to change it...
• June 16th 2007, 01:16 PM
Jhevon
Quote:

Originally Posted by sacwchiri
hello... i just started university and in my math class we are seing limits and since here it said limits well i decided to post it here....

well ive been having lots of trouble with trigonometry stuff and now they go and mix one problem with limits

Im supposed to find:

lim {cos^2(x/2)}/{cos(x)-sen(3x/2)}
x->pi

thanks for the help...

We could use L'Hopital's rule here (i don't think you would have learnt it at this time since this question was asked apparently at the begining or before your calculus class, but i can't see a way to factor things at the moment).

whenever we have the cases:

$\lim_{x \to a} \frac {f(x)}{g(x)} = \frac {0}{0} \mbox { or } \frac {\infty}{\infty}$

we can apply L'Hopital's rule (several times, if needed) to find the limit.

L'Hopital's Rule:
$\lim_{x \to a} \frac {f(x)}{g(x)} = \lim_{x \to a} \frac {f'(x)}{g'(x)}$

Let $L = \lim_{x \to \pi} \frac {\cos^2 \left( \frac {x}{2} \right)}{\cos x - \sin \left( \frac {3x}{2} \right)}$ ..........we get $\frac {0}{0}$ if we plug in $x = \pi$

Apply L'Hopital's, we get:

$L = \lim_{x \to \pi} \frac { \cos \left( \frac {x}{2} \right) \sin \left( \frac {x}{2} \right) }{ \sin x + \frac {3}{2} \cos \left( \frac {3x}{2} \right) }$ .........again, we get $\frac {0}{0}$ we we plug in $x = \pi$

Apply L'Hopital's again, we get:

$L = \lim_{x \to \pi} \frac { \frac {1}{2} \left( \cos^2 \left( \frac {x}{2} \right) - \sin^2 \left( \frac {x}{2} \right) \right)}{ \cos x - \frac {9}{4} \sin \left( \frac {3x}{2} \right)}$ ........no problems here when we plug in $x = \pi$

$\Rightarrow L = \frac {- \frac {1}{2}}{ \frac {5}{4}} = - \frac {2}{5}$
• June 16th 2007, 06:06 PM
ThePerfectHacker
I do not think L'Hopital's rule can be used here. See what happens if you substitute, or try to substitute. You get 0.
• June 16th 2007, 09:10 PM
qbkr21
Re:
Actually in order to solve this problem you must use L'Hospital's rule, and Jhevon got the correct answer.

http://item.slide.com/r/1/8/i/8JtuPv...pHQLjwd490jks/

-qbkr21
• June 17th 2007, 04:04 AM
topsquark
Quote:

Originally Posted by qbkr21
Actually in order to solve this problem you must use L'Hospital's rule, and Jhevon got the correct answer.

http://item.slide.com/r/1/8/i/8JtuPv...pHQLjwd490jks/

-qbkr21

I don't like the word "must." Just because we can't see another way to do it doesn't mean one doesn't exist!

And, after all, you can do it numerically and approach $\pi$ from both sides and at least make a case for -2/5 without doing anything fancy. Perhaps that was all that's required here for this class.

-Dan
• June 17th 2007, 09:11 AM
Jhevon
Quote:

Originally Posted by topsquark
I don't like the word "must." Just because we can't see another way to do it doesn't mean one doesn't exist!

And, after all, you can do it numerically and approach $\pi$ from both sides and at least make a case for -2/5 without doing anything fancy. Perhaps that was all that's required here for this class.

-Dan

yes, i forgot about that method...i hate it though, it's too calculation intensive and it gets annoying if you don't have a calculator that you can program the formula into
• June 17th 2007, 11:45 AM
qbkr21
Re:
Dan that is a good point however I would never go about doing a limits problem that way unless one was shooting up to - $\infty$ or $\infty$. It would be one of those limit approaches constant problems where you plug the constant in and you dealing with fraction divided by zero.

-qbkr21
• June 17th 2007, 12:00 PM
Jhevon
Quote:

Originally Posted by qbkr21
Dan that is a good point however I would never go about doing a limits problem that way unless one was shooting up to - $\infty$ or $\infty$. It would be one of those limit approaches constant problems where you plug the constant in and you dealing with fraction divided by zero.

-qbkr21

the method Dan was talking about can be used even if it is not an infinite limit. finding the limit by the numerical method means you plug in numbers increasing close to the number that causes the function to be undefined from both directions.

say you wanted to evaluate a limit as x approaches 2, but when x is 2 the function is undefined. you could plug in numbers close to 2 from the right and the left, and if the result seems to settle at a number as you get close to 2, then that is the limit, if it doesn't, then the limit does not exist. so you would try numbers like:

1.9
1.99
1.999
1.9999
1.99999
(skip 2)
2.01
2.001
2.0001
2.00001

if the result settles at a number, as we get closer to 2, then that's the limit

so for this problem, we would try, say:

3.1412
3.1413
3.1414
(skip 3.1415 ~= pi)
3.1416
3.1417
3.1418
3.15

if we do this method, we would see the result settling at -0.4 as we get close to pi. so that would be our limit. but as i said before, pluging in all those values into the formula given and calculating each result to see if they settle somewhere would be a pain. factoring algebraically would be much more fun and easier, but i couldn't see a way to do that, so i did L'Hopital's