How to find $\lim_{x \to 1} \frac{x+x^2+\sqrt{x}-3}{\sqrt[3]{x}-1}$?

Question

How to find

$$ \lim_{x \to 1} \frac{x+x^2+\sqrt{x}-3}{\sqrt[3]{x}-1}? $$

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My try : $$(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)=(x-1)$$

So we have :

$$\lim_{x \to 1} \frac{x+x^2+\sqrt{x}-3}{\sqrt[3]{x}-1} \cdot\frac{(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}{(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}=\lim_{x \to 1}\frac{(x+x^2+\sqrt{x}-3)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}{x-1}$$

Now what ?

Answer 1

Setting $$t=\sqrt[6]{x}$$ then we have $$t^6=x$$ and we get $$\frac{t^{12}+t^6+t^3-3}{t^2-1}$$

Answer 2

Hint:

$$x+x^{ 2 }+\sqrt { x } -3=x-1+x^{ 2 }-1+\sqrt { x } -1=x-1+\left( x-1 \right) \left( x+1 \right) +\sqrt { x } -1=\left( x-1 \right) \left( 1+x-1 \right) +\sqrt { x } -1=x\left( x-1 \right) +\sqrt { x } -1=\\ =x\left( \sqrt { x } -1 \right) \left( \sqrt { x } +1 \right) +\sqrt { x } -1=\left( \sqrt { x } -1 \right) \left( x\sqrt { x } +x+1 \right) $$