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How may I simplify the following so I could calculate the limit without getting $0$ in the denominator when replacing $x$ with $1$?

$$\lim_{x \to 1} \frac{\sqrt{x}-x^2}{1-\sqrt{x}}$$

Blue
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john
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3 Answers3

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

1

Let consider

$$\frac{\sqrt{x}-x^2}{1-\sqrt{x}}=\frac{\sqrt{x}-1+1-x^2}{1-\sqrt{x}}=-1+\frac{1-x^2}{1-\sqrt{x}}=-1+\frac{(1+x)(1+\sqrt{x})\color{red}{(1-\sqrt{x})}}{\color{red}{1-\sqrt{x}}}$$

and refer to

user
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  • still the denominator is 0 – john Aug 31 '20 at 08:46
  • You can cancel out the term now! – user Aug 31 '20 at 08:46
  • Is it right to distribute a fraction when it is indeterminate? – DatBoi Aug 31 '20 at 08:50
  • How you thought about adding 1... may you show the method if multiplying by 1+sqrt(x) – john Aug 31 '20 at 08:51
  • @DatBoi Yes of course, all allowed algebric manipulations before to take the limit are fine. – user Aug 31 '20 at 08:52
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    @john By this other way we obtain $$\frac{\sqrt{x}-x^2}{1-\sqrt{x}}=\frac{\sqrt{x}-x^2}{1-\sqrt{x}}\frac{1+\sqrt{x}}{1+\sqrt{x}}=\frac{(\sqrt{x}-x^2)(1+\sqrt x)}{1-x}$$ which in inconclusive again. Precisely for this reason I've proposed a different way. – user Aug 31 '20 at 08:54
  • @john Another way let $x=y^2$ $$\frac{\sqrt{x}-x^2}{1-\sqrt{x}}=\frac{y-y^4}{1-y}=\frac{y(1-y^3)}{1-y}=\frac{y(1-y)(1+y+y^2)}{1-y}$$ which reveals more clearly the standard trick used in almost all others answers. – user Aug 31 '20 at 09:07
  • @john Don't forget the old good one l'Hopital $$\lim_{x\to 1}\frac{\sqrt{x}-x^2}{1-\sqrt{x}} = \lim_{x\to 1} \frac{\frac1{2\sqrt{x}}-2x}{-\frac1{2\sqrt{x}}} $$ – user Aug 31 '20 at 09:15
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Let $r=\sqrt{x}$. Then $\sqrt{x}-x^2=r(1-r^3)=r(1-r)(1+r+r^2)$.

Ramen Nii-chan
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