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March 29, 2015

March 29, 2015

Posted by **help pls** on Thursday, May 23, 2013 at 7:42am.

x-->0

- LIMITS -
**Steve**, Thursday, May 23, 2013 at 10:48amcotx - 1/x = (x-tanx)/(x tanx)

now apply L'Hospital's Rule, so the limit is also

(1-sec^2 x)/(tanx - xsec^2 x)

and again

-2sec^2 x tanx / (sec^2 x - 2sec^2 x tanx)

= -2sec^2x tanx / (sec^2x(1-2tanx))

= 0/1

so the limit is 0

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