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

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Answer 1 · 2016-12-05T12:03:57

Rewrite so we can use #lim_(thetararr0)theta/sintheta = 1#

#cos(pi/2-x) = sinx# (by co-function identity or difference formula for cosine)

Also #cos(-theta) = cos theta# (cosine is an even function)

So #cos(x-pi/2) = cos(pi/2-x) = sinx# and

sinx = cos(x-pi/2)

#sin(pi/2-x) = cosx# (by co-function identity or difference formula for sine)

Also #sin(-theta) = -sin theta# (sine is an odd function)

So #sin(x-pi/2) = -sin(pi/2-x) = -cosx#and

cosx = -sin(x-pi/2)

Now we can rewrite

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

# = lim_(xrarrpi/2)(pi/2-x)cos(pi/2-x)/(-sin(pi/2-x))#

# = lim_(xrarrpi/2)-(pi/2-x)/(sin(pi/2-x)) * cos(pi/2-x)#

Notice that #lim_(xrarrpi/2)(pi/2-x)/(sin(pi/2-x))# is a version of #lim_(thetararr0)theta/sintheta #, so we get

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

sinx = cos(x-pi/2)