Find limit $\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}$ 
Find the following limit without using L'Hopital's rule:
  $$\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}$$

My attempt:
$$x-1=u\implies x=u+1$$
So we have:
$$\lim_{u\to 0} \dfrac{4(u+1)^2\sqrt{u+4}-17(u+1)+9}{(u)^2}$$
What now?

$$\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}=\lim_{x\to 1} \dfrac{16x^4(x+3)-(17x-9)^2}{(x-1)^2(4x^2\sqrt{x+3}+17x-9)}=$$
$$=\lim_{x\to 1} \dfrac{16x^3+80x^2+144x-81}{4x^2\sqrt{x+3}+17x-9}=\frac{159}{16}$$
