How do I solve these limits

Question

lim as x tends to pi/4 ( sinx-cosx)/(tanx-1) 

 

lim as x tends to pi/2 ( secx-tanx)

Answer 1

lim_(x→π/4)[(sinx-cosx)/(tanx-1)] has the indeterminate form 0/0

 

= lim_(x→π/4)[(sinx-cosx)/((sinx)/(cosx) - 1)]

 

= lim_(x→π/4)[(sinx - cosx)/((sinx - cosx)/(cosx))]

 

= lim_(x→π/4)[cosx] = cos(π/4) = √2/2

 

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lim_(x→π/2)(secx - tanx) = lim_(x→π/2)[i/(cosx) - (sinx)/(cosx)]

 

                                = lim_(x→π/2)[(1-sinx)/cosx]  (has 0/0 form)

 

                                = lim_(x→π/2)[(-cosx)/(-sinx)] = 0/1 = 0

                                 

                                          (By L'Hopital's Rule)