Thursday, 17 January 2013

calculus - find this limit without l'hopital's rule



I can't figure out how to get the limit in this problem. I know that 1cosxx=0 but I'm not allowed to use L'Hopital's Rule. I also already know that the answer is 2536 but I don't know the steps in between. I've already tried multiplying by the conjugates of both the numerator and the denominator but neither are getting me anywhere close. Here is the question:



limx01cos5xcos6x1



Answer



Outline: Our expression is equal to
1+cos6x1+cos5x1cos25x1cos26x,


which is
1+cos6x1+cos5xsin25xsin26x.

To find
limx0sin5xsin6x,
rewrite as
56limx0sin5x5xsin6x6x.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...