Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (anonymous):

lim as x--> 0 of (xcscx) is...

OpenStudy (anonymous):

rewrite as lim as x->0 x/sinx then use L'Hospital. Derivative of the top over derivative of the bottom. Then you get: lim x->0 1/cosx = 1

OpenStudy (anonymous):

\[\lim_{x \rightarrow 0}x \csc(x)=\lim_{x \rightarrow 0}\frac{x}{\sin(x)}=\lim_{x \rightarrow 0}\frac{1}{\cos(x)}=1\]

OpenStudy (anonymous):

What is L'Hospital's Theorem?

OpenStudy (anonymous):

http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule

OpenStudy (anonymous):

If \(\frac{f(x)}{g(x)}\) tends to a limit then \(\frac{f'(x)}{g'(x)}\) tends to the same limit.

OpenStudy (anonymous):

the limit must first be an indeterminate before you can apply L'Hopsital's rule which means that you will get that the limit becomes 0/0 or infinity/infinity then you can apply L'Hospital's

OpenStudy (anonymous):

If \(f(x)=g(x)=0 or\pm\infty\).

OpenStudy (anonymous):

i just done understand the intermediate step

OpenStudy (anonymous):

Sorry the limit of both are equal.

OpenStudy (anonymous):

dont*

OpenStudy (anonymous):

Basically, if you try to evaluate a limit and get 0/0 or \(\pm\infty/\pm\infty\), then use l'Hopital's rule.

OpenStudy (anonymous):

OH! thank you so much, I understand now

OpenStudy (anonymous):

No problem.

OpenStudy (anonymous):

l'hospital is just taking the derivative of the numerator and the denominator seperately

OpenStudy (anonymous):

Remember that this rule can also be applied to other indeterminate forms such as 1^infinity, 0^0, 0*+-infinity, infinity^0, and finally infinity - infinity. (L'Hoptial's Rule can be applied when the original limit of f(x)/g(x) results in one of these forms)

OpenStudy (anonymous):

what about something like lim as x approaches 0 of (1-cos^2(2x))/(x^2)

OpenStudy (anonymous):

We have another simple indeterminate form of 0/0 here. Just take the derivative of the numerator and the denominator separately and see if you come up with a real answer.

OpenStudy (anonymous):

if the derivative of the denominator is 2x, wont that still be 0?

OpenStudy (anonymous):

Yes. You may use L'Hospital's rule as many times as needed as long as your limit is still an indeterminate form each time. You will have a constant in the denominator after using it the second time.

OpenStudy (anonymous):

oh ok then just plug in zero?

OpenStudy (anonymous):

Yes. You should end up with 4. :)

OpenStudy (anonymous):

I got it!!!! THANK YOU

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!