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when faced with this kind of limit: $\lim _{x\to 1}\left(\frac{\sqrt[3]{x}+2\sqrt{x}-3}{x-1}\right)$

i know i can use l'hopital's rule to solve it because we get $\frac{0}{0}$ replacing $x$ with $1$, still i want to learn the other method which is cancelling the common factor, in this case i know the common factor is $x-1$

with normal polynomial function i use Euclidean division, tried it with square root but didn't seem to work.

a limit calculator gave me this answer:

$\mathrm{Cancel\:the\:common\:factors\:of}\:\frac{\sqrt[3]{x}+2\sqrt{x}-3}{x-1}$ = $\frac{2\sqrt[3]{x}+3x^{\frac{1}{6}}+3}{x^{\frac{5}{6}}+x^{\frac{2}{3}}+\sqrt{x}+\sqrt[3]{x}+x^{\frac{1}{6}}+1}$

but i don't understand yet how it is done.

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It'll be much easier to visualize if you rewrite $x^{1/6}$ as $y$

$$\frac{2y^3+y^2-3}{y^6-1}=\frac{(y-1)(2y^2+3y+3)}{(y-1)(y^5+y^4+y^3+y^2+y+1)}$$

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  • 2
    $\begingroup$ $6$ came up from the very fact that $6=$lcm$(2,3)$ $\endgroup$ – lab bhattacharjee Oct 18 '14 at 15:54

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