Hi i need help again :c

Find the following limit:
lim_(x->4) (x sqrt(x)-8)/(x-4)

(x sqrt(x)-8)/(x-4) = (x^(3/2)-8)/(x-4):
lim_(x->4) (x^(3/2)-8)/(x-4)

Factor the numerator and denominator:
lim_(x->4) ((sqrt(x)-2) (4+2 sqrt(x)+x))/((sqrt(x)-2) (2+sqrt(x)))

Cancel terms, assuming sqrt(x)-2!=0:
lim_(x->4) (4+2 sqrt(x)+x)/(2+sqrt(x))

lim_(x->4) (4+2 sqrt(x)+x)/(2+sqrt(x))  =  (4+2 sqrt(4)+4)/(2+sqrt(4))  =  3:
Answer: | =3
 

 Lim-->4 (x*sqrt(x)-8)/(x-4)  

\(\displaystyle\lim_{x\rightarrow4} \frac{ (x*\sqrt{x}-8)}{(x-4) }\\ =\displaystyle\lim_{x\rightarrow4} \frac{ (x*\sqrt{x}-8)}{(x-4) }\)

At this point I have learned from our guest.  Thankyou :)  

\(=\displaystyle\lim_{x\rightarrow4} \frac{ (x\sqrt{x}-8)}{(\sqrt{x}-2)( \sqrt{x}+2) }\\~\\~\\ \qquad \mbox{I need to get rid of the }\sqrt{x}-2\;\;\mbox{in the denominator}\\ \qquad \mbox{Now I used algebraic division to find that}\\ \qquad (x\sqrt{x}-8)\div (\sqrt{x}-2)=x+2\sqrt{x}+4\\\qquad so\\ \qquad(x\sqrt{x}-8)=(x+2\sqrt{x}+4) (\sqrt{x}-2)\\~\\~\\ =\displaystyle\lim_{x\rightarrow4} \frac{(x+2\sqrt{x}+4) (\sqrt{x}-2)}{(\sqrt{x}-2)( \sqrt{x}+2) }\\~\\ =\displaystyle\lim_{x\rightarrow4} \frac{(x+2\sqrt{x}+4) }{( \sqrt{x}+2) }\\~\\ =\frac{4+2*2+4}{2+2}\\ =3\)

Here it the graph - it has a hole in it at (4,3)

https://www.desmos.com/calculator/3jmjqdwthb