More soviet limits

Calculus Level 2

If L = lim ⁡ x → 1 x 3 − 1 x − 1 L=\displaystyle \lim_{x \rightarrow 1} \dfrac{\sqrt[3]{x}-1}{\sqrt{x}-1} , and L L can be represented in the form a b \dfrac{a}{b} , find a − b a-b . Ensure that a a and b b are relatively prime.


The answer is -1.

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Hobart Pao by Hobart Pao , 23, USA

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1 solution

If let x = u 6 x=u^{6} the limit becomes lim ⁡ u → 1 u 2 − 1 u 3 − 1 = lim ⁡ u → 1 u + 1 u 2 + u + 1 = 2 3 \displaystyle\lim_{u\to 1}\frac{u^{2}-1}{u^{3}-1}=\displaystyle\lim_{u\to 1}\frac{u+1}{u^{2}+u+1}=\frac{2}{3}

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I used the rule that lim ⁡ x → 1 x n − 1 x − 1 = n \displaystyle \lim_{x \rightarrow 1} \dfrac{x^{n}-1}{x-1} = n , easily proven by L'Hopital's rule. Then, all I had to do was substitute y = x y= \sqrt{x} . The exponent of x x in the numerator became 2 3 \dfrac{2}{3} and also the value of n.

Hobart Pao - 5 years, 8 months ago

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Interesting rule, I hadn't seen that. Thanks for giving knowledge

Hjalmar Orellana Soto - 5 years, 8 months ago

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