Question #55c75

1 Answer
Dec 14, 2017

See the following :)

Explanation:

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