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I'm often finding myself these days trying to solve tricky limits in which apparently none of the usual techniques (quoted in the title) can be applied, and I cannot find any way to go forward. Could you give me some hint?

I report two examples of that:

$1$:

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

In which: You cannot use Taylor if not on $e$ and it would not even help, cause the problematic term wouldn't turn to a polynomial. On the other hand even

$(1+\frac{1}{x})^x$

cannot be expanded using taylor.

Using the special limit:

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

brings me again to the indeterminate form $[\frac{0}{0}]$

while using De L'Hopital would not solve my indetermination cause ${\frac{d}{dx}}(f(x)^x)$ involves again $f(x)^x$ and even turning it to $e^{x\ln(f(x))}$ doesn't solve, so I'm in a blind spot here and I don't know how to go on.

How do you go on in these situations?

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    $\begingroup$ This one is not difficult. Put $1/x=t$ and so that $t\to 0$ and then the limit can be easily evaluated by a combination of standard limits and Taylor series (or L'Hospital's Rule). Express the term $(1+t)^{1/t}$ as $\exp((1/t)\log(1+t))$. The answer is $-e/2$. $\endgroup$
    – Paramanand Singh
    Commented Sep 9, 2017 at 0:19
  • $\begingroup$ The premise is not true -- you can use Taylor series (as Chappers' answer shows, or by writing directly (for $u\to 0$)$$ \frac{(1+u)^{1/u}-e}{u}=\frac{\exp(\frac{1}{u}\ln(1+u))-e}{u}=\frac{\exp(1+u/2+o(u))-e}{u}=e\frac{\exp(u/2+o(u))-1}{u}=e\frac{u/2+o(u)}{u}=\frac{e}{2}+o(1)$$ ...) what is your second example? $\endgroup$
    – Clement C.
    Commented Sep 9, 2017 at 0:32
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    $\begingroup$ @ClementC.: just a minor typo in your comment. You should have $1-u/2$ instead of $1+u/2$ and therefore the answer would be $-e/2$. $\endgroup$
    – Paramanand Singh
    Commented Sep 9, 2017 at 0:44
  • $\begingroup$ You're right. Thanks :) @ParamanandSingh $\endgroup$
    – Clement C.
    Commented Sep 9, 2017 at 0:49

3 Answers 3

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You can use Taylor series, with care: $$ (1+1/x)^x = \exp{\left(x\log{\left(1+\frac{1}{x}\right)}\right)} = \exp{\left(\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{x^{k-1}k}\right)} = ee^{-1/(2x)} e^{\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{x^{k-1}k}} \\ = e\left( 1-\frac{1}{2x} \right) + o(1/x), $$ where $o(1/x)$ means that the left-hand side minus the right-hand side divided by $1/x$ tends to zero as $x \to \infty$. Then of course some simple algebra gives the answer as $-e/2$.

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  • $\begingroup$ I love how concise this is, +1. $\endgroup$ Commented Sep 9, 2017 at 0:41
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Another method: Take the substitution $t=1/x$. Then the limit is equal to $$\lim_{t\to 0} \frac{(1+t)^{1/t}-e}{t}=\lim_{t\to 0}\frac{e^{\tfrac{\log(1+t)}{t}}-e}{t}.$$ Now as $t\to 0$, we have $\log(1+t) \in t-\frac12 t^2+O(t^3)$ hence we now consider, $$\frac{e^{1-\frac{t}2+O(t^2)}-e}{t}.$$ Now pull out an $e$ from the numerator and expand the resulting exponential into a truncated taylor series: $$e^{-\tfrac{t}2 +O(t^2)} \subset 1+(-\tfrac{t}2 +O(t^2))+O((-\tfrac{t}2 +O(t^2))^2)$$ $$\subset 1-\frac12 t +O(t^2),$$ after some simplification. Thus, all together, the expression is in $$e\frac{(1-\tfrac{t}{2}+O(t^2)-1)}{t}=e\left(-\frac12+O(t)\right),$$ and therefore, $$\lim_{t\to 0} \frac{(1+t)^{1/t}-e}{t}=-\frac{e}{2}.$$

Note: I'm assuming you are familiar with big O notation. If you are not, I am sorry. I think generally, people write this with the equal sign $f=O(g)$ with the understanding that is isn't strictly an equivalence relation like normaly equality. This is why I prefer the subset notation. Coincidentally, I asked the nearly the same question but the expression involved inverses and had limit =$-\frac1{2e}$, but this is where I learned this method.

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    $\begingroup$ This was really useful, also for the question you linked! Thanks $\endgroup$ Commented Sep 9, 2017 at 15:42
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To expand on my comment here is how I would proceed. \begin{align} L&=\lim_{t\to 0}\frac{\exp((1/t)\log(1+t))-\exp(1)}{t}\notag\\ &=\exp(1)\lim_{t\to 0}\frac{\exp((1/t)\log(1+t)-1)-1}{(1/t)\log(1+t)-1}\cdot\frac{\log(1+t)-t}{t^{2}}\notag\\ &=e\lim_{t\to 0}\frac{\log(1+t)-t}{t^{2}}\notag\\ &=-\frac{e}{2}\text{ (via Taylor or L'Hospital's Rule)} \notag \end{align} Proceeding in this fashion makes the application of Taylor series or L'Hospital's Rule very easy.

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