Question

`lim_(xrarrpi/2) 1-sinx/(x-π/2)`?

A. -2
B. -1
C. 1
D. 0
E. 2

Answer 1

If we set `u=x-pi/2` hence `x=u+pi/2` we have that

`lim (1-(sin(u+pi/2)/u))=lim(1-cosu/u)=lim((u-cosu)/u)`

Because `u->0` and `u-cosu->-1` we have that the limit is

`lim_(u->0) ((u-cosu)/u)->oo`

Hence none of the given answers is correct.

If you mean the limit `lim (1-sinx)/(x-pi/2)` then apply D'Hopital law to get

`lim_(x->pi/2) ((1-sinx)')/((x-pi/2)')=lim_(x->pi/2)(-cosx)=-cos(pi/2)=0`

Hence the correct answer is D. `0`