How do you find #lim x->pi/2# of #(sinx-1)/(x-pi/2)#?

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Answer 1 · 2016-12-02T02:09:54

#lim_(x->0) frac(sinx -1) (x-pi/2) =0#

As #sin(pi/2) =1#, this can be written as:

#lim_(x->0) frac(sinx -sin(pi/2)) (x-pi/2)#

By definition this is the value of the derivative of #sin x# for #x=pi/2#.

#lim_(x->0) frac(sinx -1) (x-pi/2) =d/dx(sinx)|_(x=pi/2) =cos(pi/2) =0#

Answer 2 · 2016-12-02T12:06:12

Please see the explanation section, below.

Form the study of trigonometry, we have

#sinx = cos(pi/2-x) = cos(x-pi/2)#.

I assume that we also have the well known trigonometric limit

#lim_(thetararr0)(1-cos theta)/theta = lim_(theta rarr 0)(cos theta -1)/theta = 0#

Now,

#(sinx-1)/((x-pi/2)) = (cos(x-pi/2)-1)/((x-pi/2))#

Ar #xrarrpi/2#, we have # lim_(theta rarr 0)(cos theta -1)/theta = 0# with #theta = x-pi/2#

#lim_(xrarrpi/2)(sinx-1)/((x-pi/2)) = lim_(xrarrpi/2)(cos(x-pi/2)-1)/((x-pi/2)) = 0#