Question #16e49

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Answer 1 · 2017-11-21T21:34:42

#lim_(x->pi/2) cosx/ln(x-pi/2+1) = -1#

Evaluate:

#lim_(x->pi/2) cosx/(ln(x-pi/2+1)#

substituting #y=x-pi/2#. Then for #x->pi/2# we have #y->0# and:

#cosx = cos(x-pi/2+pi/2) = cos(y+pi/2)#

Using the formula for the cosine of the sum of two angles:

#cos(y+pi/2) = cosycos(pi/2)-sinysin(pi/2) = -siny#

so:

#lim_(x->pi/2) cosx/ln(x-pi/2+1) = lim_(y->0) -siny/ln(1+y)#

Divide numerator and denominator by y:

#lim_(x->pi/2) cosx/ln(x-pi/2+1) = lim_(y->0) (-siny/y)/(ln(1+y)/y)#

and using the well known limits:

#lim_(t->0) sint/t = 1#

#lim_(t->0) ln(1+t)/t = 1#

we can conclude:

#lim_(x->pi/2) cosx/ln(x-pi/2+1) = -1#