How do you calculate this limit without using l’Hospital’s rule: lim 1-sinx / (π/2 - x)^2 as x → π/2 ?

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Answer 1 · 2018-04-17T20:05:21

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

Substitute:

#y = pi/4-x/2#

so that for #x->pi/2#, #y->0# and note that:

#x = pi/2-2y#

#sinx = sin(pi/2-2y) = cos2y#

so:

#lim_(x->pi/2) (1-sinx)/(pi/2-x)^2 = lim_(y->0) (1-cos2y)/(2y)^2#

Use now the trigonometric identity:

#1-cos2y = 2sin^2y#

so:

#lim_(x->pi/2) (1-sinx)/(pi/2-x)^2 = lim_(y->0) (2sin^2y)/(4y^2) = 1/2 lim_(y->0) (siny/y)^2 = 1/2#

graph{(1-sinx)/(pi/2-x)^2 [-2.5, 2.5, -1.25, 1.25]}