Question #57cc0

1 Answer
Jan 23, 2018

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