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Answer 1 · 2018-01-23T02:44:55

#lim_(x->pi/2)(pi/2-x)tan(x)->0timesoo#

Rewrite the equation as:

#lim_(x->pi/2)(pi/2-x)tan(x)=lim_(x->pi/2)((pi/2-x)sin(x))/cos(x)->0/0#

This indeterminate form can be evaluated using L'Hopital's rule.

The derivative of the numerator will be:

#d/dx(pi/2-x)sin(x)=-sin(x)+(pi/2-x)cos(x)#

(using the product rule).

The derivative of the denominator:

#d/dxcos(x)=-sin(x)#

So:

#lim_(x->pi/2)((pi/2-x)sin(x))/cos(x)#
#=lim_(x->pi/2)(-sin(x)+(pi/2-x)cos(x))/(-sin(x))=(-1)/-1=1#