Find limit $\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}$

Question

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?

Answer 1

HINT: multiply numerator and denominator by $$4x^2\sqrt{x+3}-(17x-9)$$

Answer 2

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

Answer 3

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,

$$4(1+4y+y^2)^2(y+2)-17(1+4y+y^2)+9$$

$$=(8+4y)(1+8y+16y^2+2y^2+8y^3+y^4)-17(1+4y+y^2)+9$$

$$=(8-17+9)+y(64+4-68)+y^2(8\cdot18+32-17)+O(y^3)=159y^2+O(y^3)$$