Calculate $\lim\limits_{x\to 4}\frac{2x\sqrt{x}+x-8\sqrt{x}-4}{x+\sqrt{x}-6}$ by just factoring

Question

Calculate $$\lim_{x\to 4}\dfrac{2x\sqrt{x}+x-8\sqrt{x}-4}{x+\sqrt{x}-6}$$ by just factoring

Factoring $2x\sqrt{x}+x-8\sqrt{x}-4$ gives us $(x-4)(2\sqrt{x}+1)$

$\color{red}{2x\sqrt{x}+x-8\sqrt{x}-4=x(2\sqrt{x}+1)-4(2\sqrt{x}+1)=(x-4)(2\sqrt{x}+1)}$

Factoring $x+\sqrt{x}-6$ gives us $(\sqrt{x}+3)(\sqrt{x}-2)$

Let $y=\sqrt{x}$, then $\color{blue}{y^2+y-6=(y+3)(y-2)=(\sqrt{x}+3)(\sqrt{x}-2)}$

So our limit becomes:

$$\lim_{x\to 4}\dfrac{(x-4)(2\sqrt{x}+1)}{(\sqrt{x}+3)(\sqrt{x}-2)}$$

but still this is indeterminate form, and so how would I further factor this, if posssible?

Answer 1

Let $x=t^2$ If $$x=4 $$ Then $$t=2$$

We have to find $$\lim_{t\to 2} \frac{2t^3+t^2-8t-4}{t^2+t-6} $$

$$\lim_{t\to 2} \frac{\color{blue}{(t-2)}(t+2)(2t+1)}{\color{blue}{(t-2)}(t+3)} $$

I think you can handle further.

$\color{red}{OR}$

In your equation factories as follows

$$(x-4)=(\sqrt x +2)\color{blue}{(\sqrt x-2)}$$