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I need to find $$\lim_{x\to1^-}\frac{\sqrt{1-x^2}}{\sqrt{1-x^3}}.$$ I tried l'Hôpitals, however it seems like no matter how many times you differentiate, it will still be in the indeterminate form. Is there another way to go about it?

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6 Answers 6

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Hint: Factor out $\sqrt{1-x}$. Canceling that common factor gives us $$\sqrt{\frac{(1-x)(1+x)}{(1-x)(x^2 + x + 1)}} = \sqrt{\dfrac {1+x}{x^2 + x + 1}}$$

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hint: $1-x^2 = (1-x)(1+x), 1-x^3 = (1-x)(1+x+x^2)$. You can simplify it.

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

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Use $1-x^2 = (1-x)(1+x)$ and $1-x^3 = (1-x)(1+x+x^2)$

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Hint

If $\lim_{x\to a}f(x)=L>0$ then $\lim_{x\to a}\sqrt{f(x)}=\sqrt L$.

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Hint: Factorize both the radicands in numerator and denominator, respectively as differences of squares and cubes.

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