#lim_(x->2) (x^4-4^x)/sin(pix)#
If we plug in 2 into this limit, it will become #0/0#, which is indeterminate form. And it is not okay to have this.
In this case, we have to use the L'Hospital's rule to solve this.
According to L'Hospital's rule, if #lim_(x->a) f(x)/g(x) =0/0#,
we have to differentiate the numerator and differentiate the denominator seperately and get #lim_(x->a) (f'(x))/(g'(x))#.
#lim_(x->2) (d/dx(x^4-4^x))/(d/dxsin(pix))#
#=lim_(x->2) (4x^3-d/dx4^x)/(picos(pix))#
To solve #d/dx(4^x)#, we need to know that #y=e^(lny)#.
So, #4^x=e^(ln(4^x))=e^(xln4)#
#=lim_(x->2) (4x^3-d/dx(e^(xln4)))/(picos(pix))#
#=lim_(x->2) (4x^3-ln4(e^(xln4)))/(picos(pix))#
Now, plug in 2 again, and we'll not get any indeterminate form(e.g. #0/0#, #oo/oo#...). So, we need not to use the L'Hospital rule again.
#=(4(2)^3-ln4(e^((2)ln4)))/(picos(2pi))#
#=(4*8-ln4(e^ln(4^2)))/(pi*1)#
#=(32-ln4(e^(ln16)))/pi#
#=(32-16ln4)/pi#
