Find the limit $\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x}$

Question

I am trying to evaluate the following limit

$$\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x}$$

If we use the binomial expansion of the numerator term, the answer is $\frac{1}{4}$. The same answer is obtained if we apply L'Hospital's rule.

Is it possible to solve this using something even more basic ? i.e. without recourse to binomial expansion or L'Hospital's rule or definition of derivative ? The reason I ask this is because this problem (from a textbook) comes up before the introduction of L'Hospital's rule.

user37238 wanted to know how the binomial expansion can be used.Basically, we have

$(1+x)^n= 1 + nx + \frac{n (n-1)}{2} + \frac{n (n-1)(n-2)}{2!} + \ldots$

Similarly,

$(1-x)^n= 1 - nx + \frac{n (n-1)}{2} - \frac{n (n-1)(n-2)}{2!} + \ldots$

Using these expansions with $n=\frac{1}{8}$ in the given problem gives the answer.

Could you explain the solution with binomial expansion, please. — user37238, Apr 25 '14 at 09:02
Here's another one: $$\lim_{n\to\infty}2\cdot\frac{\sqrt[n]{1+i}-\sqrt[n]{1-i}}i\cdot n$$ Enjoy! :-) — Lucian, Apr 25 '14 at 09:23
@user37238 I have edited the question and added the binomial expansion solution. — curryage, Apr 25 '14 at 09:53

Start wearing purple · Accepted Answer · 2014-04-25T09:47:03.563

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For example, use \begin{align} &\;\;\;\frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8}}{x}=\\&=\frac{1}{x}\cdot\frac{(1+x)^\frac{1}{4}-(1-x)^\frac{1}{4}}{(1+x)^\frac{1}{8}+(1-x)^\frac{1}{8}}=\\ &=\frac{1}{x}\cdot\frac{(1+x)^\frac{1}{2}-(1-x)^\frac{1}{2}}{\left[(1+x)^\frac{1}{8}+(1-x)^\frac{1}{8}\right]\left[(1+x)^\frac{1}{4}+(1-x)^\frac{1}{4}\right]}=\\ &=\frac{1}{x}\cdot\frac{(1+x)-(1-x)}{\left[(1+x)^\frac{1}{8}+(1-x)^\frac{1}{8}\right]\left[(1+x)^\frac{1}{4}+(1-x)^\frac{1}{4}\right]\left[(1+x)^\frac{1}{2}+(1-x)^\frac{1}{2}\right]}=\\ &=\frac{2}{\left[(1+x)^\frac{1}{8}+(1-x)^\frac{1}{8}\right]\left[(1+x)^\frac{1}{4}+(1-x)^\frac{1}{4}\right]\left[(1+x)^\frac{1}{2}+(1-x)^\frac{1}{2}\right]}. \end{align} Now to compute the limit, it suffices to set $x=0$ in the last expression.

edited Apr 25 '14 at 09:47

answered Apr 25 '14 at 08:52

Start wearing purple

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And setting $x=0$, where do you arrive using the last expression ? Cheers. – Claude Leibovici Apr 25 '14 at 08:58
I apologize ! I forgot the denominator meanwhile ! I just feel stupid. Cheers. – Claude Leibovici Apr 25 '14 at 09:01
@ClaudeLeibovici No problem! – Start wearing purple Apr 25 '14 at 09:02
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I think this one comes closest to OP's request for no use of derivative, LHospital, or binomial expansion. Just basic factoring is involved (+1). – coffeemath Apr 25 '14 at 09:30
@O.L. Whoa ! How is the result obtained from the last step ? The numerator is $2x$. If we divide the denominator by $x$, we somehow obtain 8 !! – curryage Apr 25 '14 at 09:36
@curryage It cancels with the numerator of the last line of my post. – Start wearing purple Apr 25 '14 at 09:38
@O.L. I don't understand. The numerator is $2x$. If we divide the denominator by $x$ and take the limit, how do we get the answer ? – curryage Apr 25 '14 at 09:43
@curryage I've updated the post, does it become more clear now? – Start wearing purple Apr 25 '14 at 09:47
@O.L. D'oh! I completely forgot the $x$ in the denominator. Apologies for the question. – curryage Apr 25 '14 at 09:52
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@O.L. I was not the only one !! – Claude Leibovici Apr 25 '14 at 10:13

score 4 · Answer 2 · answered Apr 25 '14 at 09:10

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Written in the form $2\cdot \frac{(1+x)^\frac18-(1-x)^\frac18}{2x},$ the expression is twice the difference quotient for the derivative at $t=1$ of $f(t)=t^{1/8}.$ It's actually a "symmetric" difference quotient, which for a differentiable function will approach the usual derivative. So the limit is $2\cdot f'(1)=\frac14.$

Note that then this problem could be done just from the definition of the derivative, provided one is OK with using the symmetric difference quotient, and so does not involve L'Hospital or binomial expansion.

answered Apr 25 '14 at 09:10

coffeemath

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Probably the most elegant solution here, nicely done. – orion Apr 25 '14 at 09:14
@coffeemath Nice ! but the derivative had not been introduced at this point. – curryage Apr 25 '14 at 09:25
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@orion It remains to explain how to calculate the derivatives before learning about the limits. – Start wearing purple Apr 25 '14 at 09:25

score 3 · Answer 3 · answered Apr 25 '14 at 14:51

$$\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x} =\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-1+1-(1-x)^\frac{1}{8} }{x}=\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-1 }{x}+\lim_{x \rightarrow 0} \frac{(1-x)^\frac{1}{8}-1 }{-x}=\frac{1}{8}+\frac{1}{8}=\frac{1}{4} $$

We applied the known limit for $a\in R$: $$\lim_{x \rightarrow 0} \frac{(1+x)^a-1 }{x}=a $$

score 2 · Answer 4 · answered Apr 25 '14 at 08:51

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Hint

When $x$ is small compared to $1$, $(1+x)^n \simeq 1+nx$. This is a property you must remember.

I am sure that you can take from here.

answered Apr 25 '14 at 08:51

Claude Leibovici

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Find the limit $\lim_{x \rightarrow 0} \frac{(1+x)^\frac{1}{8}-(1-x)^\frac{1}{8} }{x}$

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