# limit as x approaches (π/2), of (8cos(x))/(1-sin(x)).

Given that    lt  8cosx/1-sinx

x→π/2

lt          8cos(π/2)/1-sin(π/2)

x→π/2

lt         8* 0/(1 - 1)

x→π/2

Since numerator and denominator are zero, function is undefined, L-hospital rule

let f(x)=8cosx and g(x)=1-sinx

d/dx f(x)=d/dx 8cosx

= -8sinx

d/dx g(x)= d/dx 1-sinx

=d/dx 1 - d/dx sinx

=cosx

Rewrite the limit using L-hospital rule

lt          8cosx/1-sinx     =   lt          -8sinx/-cosx

x→π/2                              x→π/2

=      lt         - 8sin(π/2)/-cos(π/2)

x→π/2

=-2*1/0 = undefind

Since function is undefined, Apply L-hospital rule                                                                                                                                                                                                                                                                                                                                                                                                                                                                     d/dx (-8sinx)

=-8cosx

d/dx (-cosx)

=sinx

lt          -8sinx/-cosx =     lt          8cosx/sinx
x→π/2                           x→π/2

=  8cosπ/2/sinπ/2

= 8(0)/1

=0.

The solution is zero.