Question 1057730

The limit is indeterminate (0/0) so I would use L'Hospital's Rule.

L'Hospital's Rule says that Lim(x->a) { f(x)/g(x) } = Lim(x->a) {f'(x)} / Lim(x->a){g'(x)}

Yes, it means you take the derivative of the top and bottom and figure out those limits individually, then you can divide (very powerful).  It only works for 0/0,  infinity/infinity and a few select indeterminate forms.

  d(sin(x))
  ———    = cos(x)
    dx 

   d(cos(x))
   ———    =  -sin(x)
     dx 

  d(cos(2x))
  ————   =  -2sin(2x)
     dx         


So the original limit  =  Lim       { cos(x) + sin(x) } / Lim  { -2 sin(2x) }
                                     x->pi/4                           x->pi/4

                                        =  sqrt(2) / (-2)   

                                         = -0.707106