Question

Question #57cc0

Answer 1

`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`