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How to solve this? Desirable, explain in more detail. I know this should be solved by multiplying or replace a variable $$\lim\limits_{x \to 1} \dfrac{\sqrt[3]{x}-1}{\sqrt[4]{x}-1}.$$

Michael Rozenberg
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Dmitry Sokolov
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3 Answers3

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it is $$\frac{t^4-1}{t^3-1}=\frac{(t-1)(t+1)(t^2+1)}{(t-1)(t^2+t+1)}$$ use $$a^3-b^3=(a-b) \left(a^2+a b+b^2\right)$$

Dr. Sonnhard Graubner
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  • What's formula you use for numerator? – Dmitry Sokolov Sep 14 '17 at 13:40
  • I used this formula for denominator, but i dont understand how it work for numerator, if there 4n degree? – Dmitry Sokolov Sep 14 '17 at 13:49
  • Again, the substitution can only be made if we already know that the outer function is continuous: https://math.stackexchange.com/questions/167926/formal-basis-for-variable-substitution-in-limits And for that you at least need to know that the limit exists before doing the substitution. This is formally incorrect. Or am I missing something? – freakish Sep 14 '17 at 13:55
  • In our university, lecturer did not talk about such restrictions. – Dmitry Sokolov Sep 14 '17 at 14:02
  • @DmitrySokolov Well, ask him about that. Maybe he doesn't know it himself. :D In the link I've posted there's a counterexample where the substitution technique fails when the outer function is not continuous. So unless you can prove that $\lim\limits_{x \to 1} \dfrac{\sqrt[3]{x}-1}{\sqrt[4]{x}-1}$ exists (so that the function can be extended to a continuous one) before doing the substitution then the posted solution is incorrect. Or that the substitution is somehow valid in that case. – freakish Sep 14 '17 at 14:11
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$$l = \lim_{x\to1}\frac{\sqrt[3]{x}-1}{\sqrt[4]{x}-1}=\lim_{x\to1}\frac{e^{\ln{\sqrt[3]{x}}}-1}{\ln\sqrt[3]{x}}\cdot\ln\sqrt[3]{x}\cdot\frac{\ln\sqrt[4]{x}}{e^{\ln\sqrt[4]{x}}-1}\cdot\frac{1}{\ln\sqrt[4]{x}}=\lim_{x\to1}\frac{1}{3}\cdot4\cdot\frac{\ln x}{\ln x}=\frac{4}{3}$$

With $$\lim_{f(x)\to0}\frac{e^{f(x)}-1}{f(x)}=1$$

NotADeveloper
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Let $\sqrt[12]{x}=t$.

Then $$\lim_{x\rightarrow1}\frac{\sqrt[3]{x}-1}{\sqrt[4]{x}-1}=\lim_{t\rightarrow1}\frac{t^4-1}{t^3-1}=\lim_{t\rightarrow1}\frac{4t^3}{3t^2}=\frac{4}{3}$$

Michael Rozenberg
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    That doesn't seem to be formally correct. In order to do the substitution you need to know that the outer function is continuous at $1$. And for that you need to know that limit. See this: https://math.stackexchange.com/questions/167926/formal-basis-for-variable-substitution-in-limits – freakish Sep 14 '17 at 13:53