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).Originally Posted by sacwchiri
To refresh your memory:
whenever we have the cases:
$\displaystyle \lim_{x \to a} \frac {f(x)}{g(x)} = \frac {0}{0} \mbox { or } \frac {\infty}{\infty}$
we can apply L'Hopital's rule (several times, if needed) to find the limit.
L'Hopital's Rule:
$\displaystyle \lim_{x \to a} \frac {f(x)}{g(x)} = \lim_{x \to a} \frac {f'(x)}{g'(x)}$
Now, on to your question:
Let $\displaystyle L = \lim_{x \to \pi} \frac {\cos^2 \left( \frac {x}{2} \right)}{\cos x - \sin \left( \frac {3x}{2} \right)}$ ..........we get $\displaystyle \frac {0}{0}$ if we plug in $\displaystyle x = \pi$
Apply L'Hopital's, we get:
$\displaystyle L = \lim_{x \to \pi} \frac { \cos \left( \frac {x}{2} \right) \sin \left( \frac {x}{2} \right) }{ \sin x + \frac {3}{2} \cos \left( \frac {3x}{2} \right) }$ .........again, we get $\displaystyle \frac {0}{0}$ we we plug in $\displaystyle x = \pi$
Apply L'Hopital's again, we get:
$\displaystyle L = \lim_{x \to \pi} \frac { \frac {1}{2} \left( \cos^2 \left( \frac {x}{2} \right) - \sin^2 \left( \frac {x}{2} \right) \right)}{ \cos x - \frac {9}{4} \sin \left( \frac {3x}{2} \right)}$ ........no problems here when we plug in $\displaystyle x = \pi$
$\displaystyle \Rightarrow L = \frac {- \frac {1}{2}}{ \frac {5}{4}} = - \frac {2}{5}$
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!Actually in order to solve this problem you must use L'Hospital's rule, and Jhevon got the correct answer.
-qbkr21
And, after all, you can do it numerically and approach $\displaystyle \pi$ 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
Dan that is a good point however I would never go about doing a limits problem that way unless one was shooting up to -$\displaystyle \infty$ or $\displaystyle \infty$. 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
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