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Evaluate: $$ \lim_{x\to0}\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x},\ a>0,\ n\in\Bbb N $$

I've given it several tries but couldn't find an elementary method to find the limit. Two other ways that worked are L'Hospital's rule and generalized binomial expansion.


First method: $$ \lim_{x\to0} f(x) = \lim_{x\to0}\frac{g(x)}{h(x)} = \lim_{x\to0}\frac{g'(x)}{h'(x)} \\ = \lim_{x\to0}\left({1\over n}\left(a+x\right)^{{1\over n} -1} + {1\over n}(a-x)^{{1\over n} - 1}\right) = {2\over n}{a^{n-1\over n}} = \frac{2\sqrt[n]{a}}{na} $$


Second method: $$ (a+x)^{1\over n} = \sqrt[n]{a}\left(1 + {x\over a}\right)^{1\over n} =\\ \sqrt[n]{a}\left(1 + {1\over n}{x\over a} + \frac{\left({1\over n}\right)\left({1\over n} - 1\right)}{2!}\left({x\over a}\right)^2 + \cdots \right) $$

Also: $$ (a-x)^{1\over n} = \sqrt[n]{a}\left(1 - {x\over a}\right)^{1\over n} = \\ \sqrt[n]{a}\left(1 - {1\over n}{x\over a} + \frac{\left({1\over n}\right)\left({1\over n} - 1\right)}{2!}\left({x\over a}\right)^2 + \cdots \right) $$

Combining those ones may obtain: $$ \lim_{x\to0}f(x) = \lim_{x\to0}\frac{\sqrt[n]{a}\left({2x\over na} + O(x^2)\right)}{x} = \frac{2\sqrt[n]{a}}{na} $$


The problem is I'm not supposed to use derivatives for solving that limit. Also, generalized binomial expansion is somewhat too complicated as well.

Are there any elementary methods to evaluate the limit from the problem section?

I've also been trying to cast the expression to the form: $$ \lim_{x\to a}\frac{x^n - a^n}{x - a} = na^{n-1} $$

but failed.

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Just write

  • $\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x} = 2 \frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{(a+x) - (a-x)}$

Now, use $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$

So, you get

\begin{eqnarray*} \frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x} & = & 2 \frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{(a+x) - (a-x)}\\ & = & \frac{2}{\sum_{k=0}^{n-1}\sqrt[n]{(a+x)^{n-1-k}(a-x)^k}}\\ & \stackrel{x \to 0}{\longrightarrow} & \frac{2}{\sum_{k=0}^{n-1}\sqrt[n]{a^{n-1-k}a^k}} \\ & = & \frac{2}{n\sqrt[n]{a^{n-1}}} \\ & = & \frac{2\sqrt[n]{a}}{na} \\ \end{eqnarray*}

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    $\begingroup$ That is the way. Thank you for your help! $\endgroup$ – roman Apr 10 at 13:10
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    $\begingroup$ You are welcome :-) $\endgroup$ – trancelocation Apr 10 at 13:11
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First it should be easy enough to see that your limit is $$ 2 \lim_{x\to 0}\frac{\sqrt[n]{a+x}-\sqrt[n]{a}}{x} $$

Now switch variables to $y=\sqrt[n]{a+x}$, giving $x=y^n-a$: $$ \cdots = 2 \lim_{y\to b}\frac{y-b}{y^n-b^n} \qquad\text{where }b=\sqrt[n]{a}$$ Take the limit of the reciprocal instead: $$ \cdots = \frac{2}{\lim\limits_{y\to b}\frac{y^n-b^n}{y-b}} $$ and you can now use the rule that you write you already have available.

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The last formula in your question is the key here. Let $u=a+x, v=a-x$ so that both $u, v$ tend to $a$ as $x\to 0$. Also $(u-a) /x\to 1$ and $(v-a) /x\to - 1$. The given expression can be written as $$\frac{u^{1/n}-a^{1/n}}{u-a}\cdot\frac{u-a}{x}-\frac{v^{1/n}-a^{1/n}}{v-a}\cdot\frac{v-a}{x}$$ which tends to $$\frac{1}{n}a^{(1/n)-1}+\frac{1}{n}a^{(1/n)-1}=\frac{2a^{1/n}}{na}$$

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  • $\begingroup$ Nice! $(+1)$ ${}$ $\endgroup$ – Mr Pie Apr 10 at 16:43
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With $$\sqrt[n]{a+x}=u,\\\sqrt[n]{a-x}=v,$$

$$\frac{u-v}x=\frac{u^n-v^n}{x(u^{n-1}+u^{n-2}v+\cdots v^{n-1})}=\frac2{u^{n-1}+u^{n-2}v+\cdots v^{n-1}}.$$

Every term in the denominator tends to $a^{(n-1)/n}$ and there are $n$ of them.


Alternatively, let $x=at$ and pull $a$ out of the expression,

$$\frac{\sqrt[n]{a+at} - \sqrt[n]{a-at}}{at}=a^{1/n-1}\frac{\sqrt[n]{1+t} - \sqrt[n]{1-t}}{t},$$

where the fraction tends to a constant. This brings you halfway of the work.

Now,

$$\lim_{t\to0}\frac{\sqrt[n]{1+t} - 1+1-\sqrt[n]{1-t}}{t}=\left.\left(\left(\sqrt[n]{1+t}\right)'-\left(\sqrt[n]{1-t}\right)'\right)\right|_{t=0} =2\left.\left(\sqrt[n]{1+t}\right)'\right|_{t=0}.$$

You can compute the derivative or use the Taylor development of$\sqrt[n]{1+t}$, and establish that the constant is $\dfrac2n$.

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Alternatively: $$\lim_{x\to0}\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x}= \lim_{x\to0}\frac{\sqrt[n]{a-x}\cdot \left(\sqrt[n]{\frac{a+x}{a-x}}-1\right)}{x}\stackrel{\frac{a+x}{a-x}=t^n}{=}\\ \sqrt[n]{a}\cdot \lim_{t\to1}\frac{t-1}{\frac{a(t^n-1)}{t+1}}= \frac{2\sqrt[n]{a}}{a}\cdot\lim_{t\to1}\frac{ t-1}{t^n-1}=\frac{2\sqrt[n]{a}}{an}.$$

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  • $\begingroup$ Incidentally, I will definitely use that shorthand notation; quick and easy! $(+1)$ $\endgroup$ – Mr Pie Apr 10 at 16:46
  • $\begingroup$ I am glad it was helpful. Good luck. $\endgroup$ – farruhota Apr 10 at 16:51
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Rewrite the expression as $$\frac{\sqrt[n]{a+x} - \sqrt[n]{a-x}}{x}=\frac{\sqrt[n]{a+x} - \sqrt[n]{a}}{x}+\frac{\sqrt[n]{a-x} - \sqrt[n]{a}}{-x}$$ and observe each fraction is a rate of variation from $a$. hence the limit of each is the derivative at $a$.

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    $\begingroup$ The OP writes explicitly that he's looking for a way without derivatives. $\endgroup$ – Henning Makholm Apr 10 at 13:42

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