How do you find the limit of # (x-pi/2)tanx# x as it approaches pi/2?

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Answer 1 · 2015-09-25T01:35:33

Use #cosx = -sin(x-pi/2)# and the fundamental trigonometric limit.

#(x-pi/2)tanx = (x-pi/2)sinx/cosx#.

Since #lim_(xrarrpi/2)(x-pi/2) = 0#

we should consider trying to use #lim_(thetararr0)sintheta/theta = lim_(thetararr0)theta/sintheta = 1# (With #theta = x-pi/2#)

To do this we need #sin(x-pi/2)#

Using either the sum formula or using #cosx = sin(pi/2-x)# and the fact that sine is an odd function, we get:

#cosx = -sin(x-pi/2)#.

So,

#lim_(xrarr pi/2)(x-pi/2)tanx = lim_(xrarr pi/2)(x-pi/2)sinx/cosx#

# = lim_(xrarr pi/2)(x-pi/2)sinx/(-sin(x-pi/2)#

# = lim_(xrarr pi/2)-(x-pi/2)/sin(x-pi/2) * sinx#

# = -(1)(1) = -1#