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i have to solve $ \lim_{x\to 0} \frac{{(1+x})^{1/x}-e+ex/2}{x^2}$ and in the hint it is advised to use the algebraic expansion of $(1+x)^{1/x}$.

Also i know that the algebraic expansion of $(1+x)^{m} = 1+mx+m(m-1){x^2}/2+...$ but this is in the case where m is an integer. So please help me here with the limit.

Parul
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3 Answers3

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Note that you can develop $$ \begin{gathered} \left( {1 + x} \right)^{\,1/x} = \exp \left( {\frac{1} {x}\ln \left( {1 + x} \right)} \right) = \exp \left( {1 - \frac{x} {2} + \frac{{x^{\,2} }} {3} + O\left( {x^{\,3} } \right)} \right) = \hfill \\ = \exp \left( 1 \right)\exp \left( { - \frac{x} {2}} \right)\exp \left( {\frac{{x^{\,2} }} {3}} \right)\exp \left( {O\left( {x^{\,3} } \right)} \right) = \hfill \\ = e\left( {1 - \frac{x} {2} + \frac{{x^{\,2} }} {8}} \right)\left( {1 + \frac{{x^{\,2} }} {3}} \right) + O\left( {x^{\,3} } \right) = e\left( {1 - \frac{x} {2} + \frac{{11\;x^{\,2} }} {{24}} + O\left( {x^{\,3} } \right)} \right) \hfill \\ \end{gathered} $$ from which $$ \begin{gathered} \frac{1} {{x^{\,2} }}\left( {\left( {1 + x} \right)^{\,1/x} - \left( {1 - x/2} \right)e} \right) = \hfill \\ = \frac{1} {{x^{\,2} }}\left( {e\left( {1 - \frac{x} {2} + \frac{{11\;x^{\,2} }} {{24}}} \right) - \left( {1 - x/2} \right)e + O\left( {x^{\,3} } \right)} \right) = \hfill \\ = \frac{{11}} {{24}}e + O\left( x \right) \hfill \\ \end{gathered} $$

Botond
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G Cab
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Hint: Use $$ f(x)=f(0)+f'(0)x+\frac{1}{2}f''(c)x^2 $$ where $c$ is between $0$ and $x$.

Update: Let $$ f(x)=（1+x)^{1/x}. $$ and then $$ f'(x)=-（1+x)^{1/x}\frac{x-\ln(1+x)}{x^2(1+x)}, f''(x)=\frac{(x+1)^{\frac{1}{x}-2}(-(3 x+1) x^2+2 (x+1) x^2 \ln (x+1)+(x+1)^2\ln ^2(x+1))}{x^4}. $$ Clearly \begin{eqnarray} f(0)&=&\lim_{x\to0}（1+x)^{1/x}=e,f'(0)=\lim_{x\to0}f'(x)=-\frac{e}{2},f''(0)=\frac{11e}{12}. \end{eqnarray} So $$ \lim_{x\to 0} \frac{{(1+x})^{1/x}-e+ex/2}{x^2}=\lim_{x\to0}\frac{\frac{1}{2}f''(c)x^2}{x^2}=\frac{1}{2}f''(0)=\frac{11e}{24}.$$

xpaul
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    isn't it suppose to be f prime in the second term, also how is it useful here? – Parul Oct 18 '16 at 14:16
  • @Parul, yes, you are right. – xpaul Oct 18 '16 at 14:44
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    You implicitly assume that the function and its derivatives are continuous at $0$. The given function has a removable discontinuity at $0$ and even after removing it, it's difficult to prove that the derivatives are continuous at $0$. – Paramanand Singh Dec 28 '19 at 03:24
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As Zelos points out, you may compute the limit as $x$ approaches 0 by considering the value of your expression at $x=1/m$ for $m$ any nonzero integer. Then $1/x$ is an integer, and your formula applies.

Alternatively your formula, the binomial theorem, applies even when $1/x$ is not an integer, with the proviso that the series will not terminate for negative or non-integer exponents.

Hence you may even use your formula

$$(1+x)^{1/x} = 1 + x/x+\dotsb$$

with or without the assumption that $1/x$ is an integer.

In more detail, the binomial theorem gives

$$(1+x)^{1/x} = \sum_{m=0}^\infty \frac{1}{m!}\frac{1}{x}\cdot\left(\frac{1}{x}-1\right)\cdot\dotsb\cdot\left(\frac{1}{x}-m+1\right)x^m $$

Let us expand the expression inside the summation to collect the terms of degree 0, 1, and 2 in $x$.

To find the terms of the expanded product, you must choose one term from each binomial factor in the product. The degree of the term will be determined by how many times you choose the $1/x$ term.

The only way to get a degree 0 term is to choose the $1/x$ from each of the $m$ binomial factors in parentheses, so you get $\frac{1}{m!}\frac{x^m}{x^m}=\frac{1}{m!}$.

If you choose the second term in the binomial factor exactly once, and the $1/x$ $m-1$ times, then you get $m$ linear terms. Those the linear coefficient is the sum of numbers from 0 to $m-1$.

To get a quadratic term, you need to choose the second term exactly twice. The coefficient is thus the sum of all products of two numbers from 0 to $m-1$.

In summary

$$ \frac{1}{m!}\frac{1}{x}\left(\frac{1}{x}-1\right)\cdot\dotsb\cdot\left(\frac{1}{x}-m+1\right)x^m = \frac{1}{m!} - \frac{1}{m!}\left(\sum_{j=0}^{m-1} j\right)x + \frac{1}{m!}\left(\sum_{j=0}^{m-1} \sum_{k=j+1}^{m-1} jk\right)x^2 +o(x^3)$$

Using the Faulhaber identities to evaluate those summations, we get

$$ \frac{1}{m!} - \frac{1}{m!}\left(\frac{m(m-1)}{2}\right )x + \frac{1}{m!}\left(\frac{m(m-1)(m-2)(3m-1)}{24}\right )x^2 +o(x^3) $$

Now summing each term over $m$ we see the constant term is $\sum\frac{1}{m!}=e$, the linear coefficient is $-\frac{1}{2}\sum\frac{1}{(m-2)!}=-e/2$, and the quadratic coefficient is (after some reindexing) $\frac{1}{24}\sum\frac{3m-1}{(m-3)!} = \frac{11e}{24}$, which gives the answer.

ziggurism
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