We have:
#lim_(x->pi/2)(cos(x)*sec(5x))#
Using the fact that #secx=1/cosx#, we have:
#lim_(x->pi/2)(cos(x)*1/cos(5x))#
=>#lim_(x->pi/2)(cos(x)/cos(5x))#
When plugging #pi/2# in the place of #x#, we get #0/0#
We use L'Hospital's Rule, which states that #lim_(x->c)(f(x))/(g(x))=(f'(c))/(g'(c))# after we get an indeterminate form after substituting #c# in the place of #x#.
Now, remember that:
#d/dx(cosx)=-sinx#
We also have that:
#d/dx(cos5x)=-5sin5x# using the chain rule.
Chain rule: When #f(x)=g(h(x))#, then #f'(x)=g'(h(x))*h'(x)#
Therefore, #lim_(x->pi/2)(cos(x)/cos(5x))=(-sin(pi/2))/(-5sin(5*pi/2))#
Computing this, we get: #lim_(x->pi/2)(cos(x)/cos(5x))=0.2#
