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I cannot figure out how to get around the zero numerator and denominator in order to compute the limit below:

$$\lim_{x \to 1}\frac{\left(x^\frac{1}{5}\right)-1}{ \left( x^\frac{1}{6}\right) -1}$$

I tried:

$$ \lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{1}{6} - 1) (x^\frac{1}{6} + 1) } $$

$$\lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{2}{6} - 1) } $$

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    $\begingroup$ Is L'Hôpital's rule allowed? $\endgroup$ Commented Jan 28, 2018 at 13:37
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    $\begingroup$ Substitute $$x=t^{30}$$ $\endgroup$ Commented Jan 28, 2018 at 13:37
  • $\begingroup$ Please, if you are ok, you can accept the answer and set it as solved. Thanks! $\endgroup$
    – user
    Commented Jan 31, 2018 at 22:28

9 Answers 9

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Just to avoid L' Hospital's rule, consider the following: $$\frac{x^{1/5}-1}{x^{1/6}-1}=\frac{x^{6/30}-1}{x^{5/30}-1}=\frac{(x^{1/30})^6-1}{(x^{1/30})^5-1}=\frac{(x^{1/30}-1)(x^{5/30}+x^{4/30}+\dots+1)}{(x^{1/30}-1)(x^{4/30}+x^{3/30}+\dots+1)}$$ So: $$\lim_{x\to1}\frac{x^{1/5}-1}{x^{1/6}-1}=\lim_{x\to1}\frac{(x^{1/30}-1)(x^{5/30}+x^{4/30}+\dots+1)}{(x^{1/30}-1)(x^{4/30}+x^{3/30}+\dots+1)}=\lim_{x\to1}\frac{x^{5/30}+x^{4/30}+\dots+1}{x^{4/30}+x^{3/30}+\dots+1}=\frac{6}{5}$$

Hope this provided an alternative! :)

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  • $\begingroup$ In the last row, the $1$ is in the wrong place (It's in the exponent), but I can't fix it. $\endgroup$
    – Botond
    Commented Jan 28, 2018 at 13:49
  • $\begingroup$ Hehe, you' re right! I fixed it! Thank you! :) $\endgroup$ Commented Jan 28, 2018 at 13:50
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Here's another solution depending on what you're allowed to use: substitute $u=x^{\frac{1}{6}}$.

Then we get $$\lim_{u\rightarrow 1}\frac{u^{\frac{6}{5}}-1}{u-1}$$ and we can recognise this as the derivative of $u\mapsto u^{\frac{6}{5}}$ at 1, which is of course $\frac{6}{5}$.

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  • $\begingroup$ Very nice solution! $\endgroup$
    – user
    Commented Jan 29, 2018 at 8:18
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    $\begingroup$ @gimusi: $f(x)/g(x)\to f(g^{-1}(u))/u\to(f(g^{-1}(u)))'$ is L'Hospital in disguise. $\endgroup$
    – user65203
    Commented Jan 29, 2018 at 8:22
  • $\begingroup$ It was allowed in the OP. Don’t you like it? $\endgroup$
    – user
    Commented Jan 29, 2018 at 8:31
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By L'Hopital's rule:\begin{align}\lim_{x\to1}\frac{x^\frac15-1}{x^\frac16-1}&=\lim_{x\to1}\frac{\frac15x^{-\frac45}}{\frac16x^{-\frac56}}\\&=\frac65.\end{align}

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By the L'Hospital's rule we obtain: $$\lim_{x\rightarrow1}\frac{x^{\frac{1}{5}}-1}{x^{\frac{1}{6}}-1}=\lim_{x\rightarrow1}\frac{\frac{1}{5}x^{-\frac{4}{5}}}{\frac{1}{6}x^{-\frac{5}{6}}}=\frac{\frac{1}{5}}{\frac{1}{6}}=\frac{6}{5}$$

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Let $x=1+y$ with $y\to0$ and use

$$(1+y)^a=1+ay+o(y)$$

$$\frac{\left(x^\frac{1}{5}\right)-1}{ \left( x^\frac{1}{6}\right) -1}=\frac{1+\frac{1}{5}y-1+o(y)}{1+ \frac{1}{6}y -1+o(y)}=\frac{\frac{1}{5}y+o(y)}{\frac{1}{6}y+o(y)}=\frac{\frac{1}{5}+o(1)}{\frac{1}{6}+o(1)}\to\frac65$$

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Correct me if wrong:

Let $y^{30}:=x$.

Then: $ x^{1/5} = y^6 $; $x^{1/6} =y^5$.

Numerator:

$n(y)=(y^6-1) =$

$(y-1)(y^5+y^4+y^3+y^2+y+1);$

Denominator:

$d(y) =(y^5-1)= $

$(y-1)(y^4+y^3+y^2+y+1).$

$\lim_{y \rightarrow 1}\dfrac{n(y)}{d(y)}=$

$\lim_{y \rightarrow 1}\dfrac{y^5 +y^4+y^3+y^2+y+1}{y^4+y^3+y^2+y+1}=6/5.$

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Hint:

As lcm$(5,6)=30,$ choose $x^{1/30}=y\implies x^{1/5}=y^6, x^{1/6}=?$

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Take $u=x^{\frac{1}{5}}$ and $v=x^{\frac{1}{6}}$.

$\displaystyle u-1=\frac{u^5-1}{u^4+u^3+u^2+u+1}=\frac{x-1}{u^4+u^3+u^2+u+1}$

$\displaystyle v-1=\frac{v^6-1}{v^5+v^4+v^3+v^2+v+1}=\frac{x-1}{v^5+v^4+v^3+v^2+v+1}$

$\displaystyle \frac{u-1}{v-1}=\frac{v^5+v^4+v^3+v^2+v+1}{u^4+u^3+u^2+u+1}\to\frac{6}{5}$ as $x\to 1$

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By the generalized binomial theorem,

$$(1+t)^\alpha-1=\alpha t+\frac{\alpha(\alpha-1)}2t^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}t^3+\cdots$$ and higher order terms.

Hence your limit is essentially

$$\lim_{t\to0}\frac{\dfrac t5}{\dfrac t6}.$$

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    $\begingroup$ Nice ante Taylor derivation! $\endgroup$
    – user
    Commented Jan 29, 2018 at 8:32

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