First slide
Continuity
Question

 If f(x)=(cosx)1sinx     for x0k     for x=0 then value of k, so that f is differentiable at x=0 is 

Moderate
Solution

Ltx0  cosx1sinx=k

k=eLtx01sinxlogcosx=eLtx0  tanxcosx

=e0=1

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