lim as x--> 0 of (xcscx) is...
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
\[\lim_{x \rightarrow 0}x \csc(x)=\lim_{x \rightarrow 0}\frac{x}{\sin(x)}=\lim_{x \rightarrow 0}\frac{1}{\cos(x)}=1\]
What is L'Hospital's Theorem?
If \(\frac{f(x)}{g(x)}\) tends to a limit then \(\frac{f'(x)}{g'(x)}\) tends to the same limit.
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
If \(f(x)=g(x)=0 or\pm\infty\).
i just done understand the intermediate step
Sorry the limit of both are equal.
dont*
Basically, if you try to evaluate a limit and get 0/0 or \(\pm\infty/\pm\infty\), then use l'Hopital's rule.
OH! thank you so much, I understand now
No problem.
l'hospital is just taking the derivative of the numerator and the denominator seperately
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)
what about something like lim as x approaches 0 of (1-cos^2(2x))/(x^2)
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.
if the derivative of the denominator is 2x, wont that still be 0?
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.
oh ok then just plug in zero?
Yes. You should end up with 4. :)
I got it!!!! THANK YOU
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