Sunday, 15 December 2013

trigonometry - Limit and L'hopital's Rule



How do you evaluate the following limit?

limx0xsin1xxsinx


When I is L'Hopital's rule twice, I get:
limx0(x2+2)cscx(1x2)3/2

Which doesn't exits. If the limit DNE then can't use L'Hopital's rule.



So, how do I find this limit?


Answer



limx0xsin1xxsinx=limx0sin1xxlimx01xsinxx3limx01x



The first & the last limits are elementary and




for the second use Are all limits solvable without L'Hôpital Rule or Series Expansion


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