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I have done it thus far:

$$\lim_{x \to 0}\frac{{(x+1)}^\frac{1}{x}-e}{x} = \bigg[\frac{0}{0}\bigg] = \frac{((x+1)^\frac{1}{x}-e)'}{x'}=({x+1})^\frac{1}{x} \cdot \left(\frac{\ln(x+1)}{x}\right)' = \\({x+1})^\frac{1}{x} \cdot \frac{\left(\frac{x}{x+1}-(\ln(x+1)\right)}{x^2} = ({x+1})^\frac{1}{x} \cdot \left(\frac{\frac{1}{x+1}}{x} - \frac{\ln(x+1)}{x^2}\right) $$

I don't know what to do next.

Also could someone elaborate as to why when I have to find a derivative for $\frac{\ln(x+1)}{x}$ I need to use the quotient rule, but when I first derived the fraction that is given I could derive numerator and denominator separately?

Michael Rybkin
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user
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    In regards to your last question, you are applying something called L'Hopital's rule if you differentiate numerator and denominator separately. This has nothing really to do with quotient rule, it helps you evaluate limits where your expression is indeterminate. – psa May 06 '19 at 21:32
  • Two errors. 1. The indeterminate form is 0/0 not $\infty/\infty$ 2. No square in denominator – Fomalhaut May 06 '19 at 21:35
  • @ErotemeObelus Yes, but the OP has correctly differentiated it. – Peter Foreman May 06 '19 at 21:35
  • @PeterForeman my phone's screen is tiny so I didn't see the multiplication. – Fomalhaut May 06 '19 at 21:37

4 Answers4

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$$\begin{align} \lim_{x\to0}\left(\frac{\frac{x}{x+1}-\ln{(x+1)}}{x^2}\right) &=\lim_{x\to0}\left(\frac{x-(x+1)(x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4))}{x^2(x+1)}\right)\\ &=\lim_{x\to0}\left(\frac{-x^2+\frac{x^2}{2}+\frac{x^3}{2}-\frac{x^3}{3}+O(x^4)}{x^3+x^2}\right)\\ &=\lim_{x\to0}\left(\frac{-\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)}{x^3+x^2}\right)\\ &=\lim_{x\to0}\left(\frac{-x+\frac{x^2}{2}+O(x^3)}{3x^2+2x}\right)\\ &=\lim_{x\to0}\left(\frac{-1+x+O(x^2)}{6x+2}\right)\\ &=-\frac12\\ \end{align}$$ Hence the initial limit becomes $$\overbrace{\lim_{x\to0}\left((x+1)^\frac1x\right)}^{\large{\to\, e}}\cdot\overbrace{\lim_{x\to0}\left(\frac{\frac{x}{x+1}-\ln{(x+1)}}{x^2}\right)}^{\large{\to\, -\frac12}}=-\frac{e}{2}$$

Peter Foreman
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So, we apply L'Hôpital's rule, and get $((1+x)^{\dfrac 1x})'=(e^{\ln(1+x)^{\dfrac 1x}})'=(e^{\ln(1+x)^{\dfrac1x}}\cdot(\dfrac 1x\ln(1+x))'=(1+x)^{\dfrac 1x}\cdot (-\dfrac 1{x^2}\ln(1+x)+\dfrac1x\dfrac 1{1+x})$.

Now the first term converges to $e$; and the quantity in parentheses converges easily, upon a couple more applications of L'Hôpital to $-\dfrac 12$.

So we get $-\dfrac e2$.

  • How can you apply L'Hospital's rule if it's [ e * 0 ] ? – user May 07 '19 at 11:10
  • The quantity in parentheses is of the form $\dfrac00$ after you combine it into a single fraction. –  May 07 '19 at 15:51
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You want to compute the derivative of $$ f(x)=(1+x)^{1/x} $$ Consider $\log f(x)=\dfrac{\log(1+x)}{x}$ and $$ \frac{f'(x)}{f(x)}=\frac{\dfrac{x}{1+x}-\log(1+x)}{x^2} $$ so you have $$ f'(x)=f(x)\frac{x-(1+x)\log(1+x)}{x^2(1+x)} $$ Note also that $\lim\limits_{x\to0}\dfrac{f(x)}{1+x}=e$, so you need to compute $$ \lim_{x\to0}\frac{x-(1+x)\log(1+x)}{x^2} $$ With a further application of l'Hôpital, $$ \lim_{x\to0}\frac{1-1-\log(1+x)}{2x}=-\frac{1}{2}\lim_{x\to0}\frac{\log(1+x)}{x}=-\frac{1}{2} $$ Reinserting the factor $e$, the sought limit is $-e/2$.

egreg
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With binomial expansion $$\begin{align} (1+x)^{\frac1x}-e &=1+\frac1x\frac{x}{1!}+\frac1x(\frac1x-1)\frac{x^2}{2!}+\frac1x(\frac1x-1)(\frac1x-2)\frac{x^3}{3!}+\cdots-\left(1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+\cdots\right)\\ \frac{(1+x)^{\frac1x}-e }{x}&= -\dfrac{1}{2!}-\dfrac{3-2x}{3!}-\dfrac{6-11x+6x^2}{4!}+\cdots\\ &\to-\dfrac{1}{2!}\left(1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots\right)\hspace{1cm}\text{as}\hspace{0.3cm} x\to0.\\ &=-\dfrac12e \end{align}$$

Nosrati
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