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$$ \lim_{x \to \infty} \frac{(x+1)^{x+1}}{x^x} - e\cdot x $$ $$ =\lim_{x \to \infty} \frac{(x+1)^{x}}{x^x} \cdot (x+1) - e\cdot x $$ $$ =\lim_{x \to \infty} e\cdot (x+1) - e\cdot x $$ $$ =\lim_{x \to \infty} e $$ $$ =e $$ But when I plot the function or ask wolfram alpha, it shows e/2. I'm assuming that simplifying $\frac{(x+1)^x}{x^x}$ is causing this discrepancy, but I can't figure out why.

EDIT

I understand now that I can't simplify the limits because the second goes to infinity. If someone could provide a proof for e/2, I would be very grateful!

throwaway toa
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    $\lim f(x)g(x)=\lim f(x) \lim g(x)$ when both those individual limits exist. In your example, this is not the case, since $\lim (x+1)$ is infinite. – Malady Jun 28 '24 at 00:58
  • You made up a rule, so it's hard to tell if the current criticisms in the comment and answer are applicable. Seemingly, you are asking why $\lim_{x\to\infty}f(x)/x=L$ doesn't imply $\lim_{x\to\infty}(f(x)-Lx)=0$. The answer is that going from one to the other gives an indeterminate form $0\cdot\infty$, so yes, it has something to do with infinity, but when you just evaluate a limit like you've done, it's not a question of "why didn't this work?". It's, turning it around, "why do you think it should have?" – Brian Moehring Jun 28 '24 at 01:39
  • If there is an answer you like, you can accept the answer. – Lucenaposition Jun 29 '24 at 00:29

6 Answers6

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\begin{align} &\lim_{x \to \infty} \frac{(x+1)^{x}}{x^x} \cdot (x+1) - e\cdot x\\ =&\lim_{x \to \infty} e\cdot (x+1) - e\cdot x\\ \end{align} But when I plot the function or ask wolfram alpha, it shows e/2. I'm assuming that simplifying $\frac{(x+1)^x}{x^x}$ is causing this discrepancy, but I can't figure out why.

$$\lim_{x\to\infty}f(x)\cdot g(x)=\lim_{x\to\infty}f(x)\cdot\lim_{x\to\infty}g(x)$$ when both limits are real numbers.
You used $$\lim_{x\to\infty}\frac{(x+1)^x}{x^x}\cdot(x+1)=\lim_{x\to\infty}\frac{(x+1)^x}{x^x}\cdot\lim_{x\to\infty}(x+1)$$ but the second limit is infinite.

Lucenaposition
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$$\lim_{x \to \infty} \frac{(x+1)^{x+1}}{x^x} - e\cdot x = \lim_{x \to \infty}( \frac{(x+1)^{x+1}}{x^{x+1}} - e)\cdot x =$$

now $u = \frac 1x$

$$ = \lim_{u \to 0^+}\frac{( (u+1)^{\frac1u+1} - e)}{u}= \lim_{u \to 0^+}\frac{( (u+1)^{\frac1u}(u+1) - e)}{u} =$$

$$= e + \lim_{u \to 0^+}\frac{( (u+1)^{\frac1u} - e)}{u}$$

now we do the l'Hopital rule

$$\lim_{u \to 0^+}\frac{( (u+1)^{\frac1u} - e)}{u} = \lim_{u \to 0^+} ((u+1)^{\frac1u})'$$

$f(x) = (x+1)^{\frac1x} \implies \ln f = \frac1x \ln (x+1) \implies \frac{f'}f = -\frac1{x^2} \ln(x+1) + \frac1{x(x+1)}$. Call our desired limit $y$:

$$\frac ye = \lim_{u \to 0^+} (\frac1{u(u+1)} - \frac{\ln(u+1)}{u^2}) = \lim_{u \to 0^+} \frac{(\frac u{(u+1)} - \ln(u+1))}{u^2} = \lim_{u \to 0^+} \frac{\frac1{(u+1)^2}- \frac1{u+1}}{2u} =$$

$$= \lim_{u \to 0^+} \frac{\frac1{(u+1)}- 1}{2u(u+1)} = \lim_{u \to 0^+} \frac{\frac{-u}{(u+1)}}{2u(u+1)} = - \frac12$$

so $y = -\frac e2$ and our limit is $e-\frac e2 = \frac e2$

hellofriends
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So the limit you have is

$$ \lim_{x\to\infty}\left[ \Big(1+\frac{1}{x}\Big)^x(x+1)-ex \right] $$

You replaced $\big(1+\frac{1}{x}\big)^x$ with $e$. Is that valid? Well, the difference between $e$ and $\big(1+\frac{1}{x}\big)^x$ does tend to $0$, but this difference is effectively multiplied by $(x+1)$, and that product does not tend to $0$. We can do the ol' "add and subtract the same thing" trick (in this case, $e$) to get

$$ \Big(\Big(1+\frac{1}{x}\Big)^x-e+e\Big)(x+1)-ex ~=~\Big(\Big(1+\frac{1}{x}\Big)^x-e\Big)(x+1)+e. $$

From the Newton-Mercator expansion we have

$$ \Big(1+\frac{1}{x}\Big)^x=\exp\Big(x\ln\Big(1+\frac{1}{x}\Big)\Big)=\exp\Big(x\Big(\frac{1}{x}-\frac{1}{2x^2}+\cdots\Big)\Big) $$

$$ =\exp\Big(1-\frac{1}{2x}+\cdots\Big)=e\exp\Big(-\frac{1}{2x}+\cdots\Big)=e\Big(1-\frac{1}{2x}+\cdots\Big) $$

Therefore, what you want to take the limit of is

$$ \Big(e\Big(1-\frac{1}{2x}+\cdots\Big)-e\Big)(x+1)+e $$

This becomes $e/2$ in the limit (exercise). Note the higher order terms ("$\cdots$") tend to $0$.

coiso
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$$A=\frac{(x+1)^{x+1}}{x^x}$$ $$ \log(A)=(x+1)\log(x+1)-x \log(x)$$ By Taylor $$\log(A)=(\log (x)+1)+\frac{1}{2 x}-\frac{1}{6 x^2}+O\left(\frac{1}{x^3}\right)$$ By Taylor again $$A=e^{\log(A)}=e x+\frac{e}{2}-\frac{e}{24x}+O\left(\frac{1}{x^2}\right)$$

Claude Leibovici
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The mistake you made is that you replaced the limit with a finite value $e$ even if the limit of the product $ex$ is infinite. This led you to the incorrect value $e$. The correct limit is $\frac e2$. It can be shown as follows.

$$\begin{aligned}(x+1)^{x+1}& = \exp\left((1+x) \ln(1+x)\right) \\&= \exp\left((1+x)\left(\ln x + \frac{1}{x} - \frac{1}{2x^2} + O\left(\frac1{x^3}\right)\right)\right)\\ &= \exp\left( \ln x+ \frac1x +x\ln x+1- \frac{1}{2x} + O\left(\frac1{x^2}\right)\right) \\ &= x^{x+1}e \left(1 + \frac{1}{2x}+ O\left(\frac1{x^2}\right)\right)\end{aligned}$$

Now, $$\dfrac{(x+1)^{x+1}}{x^x} = ex + \frac{e}{2} + O\left(\frac1x\right)\implies\color{red}{\dfrac{(x+1)^{x+1}}{x^x} -ex = \frac{e}{2} + O\left(\frac1x\right)}$$

which clearly shows that the limit is $\frac e2$ as $x\to\infty$.

Hence, $$\color{blue}{\boxed{~\lim_{x \to \infty}\left( \frac{(x+1)^{x+1}}{x^x} - e\cdot x\right)=\frac e2~}}$$

I hope this helps!

Pustam Raut
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Your mistake was already pointed out, as an alternative we can proceed as follows

$$\frac{(x+1)^{x+1}}{x^x} - e\cdot x=x\cdot e^{(x+1)\log\left(1+\frac1x\right)}-e\cdot x=$$

$$e\cdot\frac{e^{(x+1)\log\left(1+\frac1x\right)-1}-1}{(x+1)\log\left(1+\frac1x\right)-1}\cdot\left(x(x+1)\log\left(1+\frac1x\right)-x\right)\to e\cdot 1 \cdot \frac12=\frac e 2$$

indeed by standard limt

$$(x+1)\log\left(1+\frac1x\right)-1 =\left(1+\frac1x\right)\frac{\log\left(1+\frac1x\right)}{\frac1x} -1\to1\cdot 1-1= 0$$

and using that result

$$x(x+1)\log\left(1+\frac1x\right)-x=x^2\log\left(1+\frac1x\right)-x+x\log\left(1+\frac1x\right)=$$ $$=\frac{\log\left(1+\frac1x\right)-\frac1x}{\frac 1{x^2}}+\frac{\log\left(1+\frac1x\right)}{\frac1x}\to -\frac12+1=\frac12$$

user
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