Question #57cc0

Question

905 views around the world

You can reuse this answer
Creative Commons License

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:

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