<s>[INST] 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?
 [/INST] Let $\sqrt{x+3}-2=y\implies x=y^2+4y+1$
$$\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}=\lim_{y\to0}\dfrac{4(1+4y+y^2)^2(y+2)-17(1+4y+y^2)+9}{y^2(y+4)^2}$$
Now,
$$4(1+4y+y^2)^2(y+2)-17(1+4y+y^2)+9$$
$$=(8+4y)(1+8y+16y^2+2y^2+8y^3+y^4)-17(1+4y+y^2)+9$$
$$=(8-17+9)+y(64+4-68)+y^2(8\cdot18+32-17)+O(y^3)=159y^2+O(y^3)$$
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