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Find the limit. Use l'Hospital's Rule if appropriate.

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

asked Dec 25, 2012 in PRECALCULUS by futai Apprentice

2 Answers

0 votes
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.
answered Dec 25, 2012 by friend Mentor
0 votes

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

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Since function is undefined, apply L-hospital rule.

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The solution is 0.

answered Jun 6, 2014 by joly Scholar

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