# How to behave in limits calculus when both special limits, taylor and De l'Hopital cannot be applied?

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?

• 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 $-e/2$. Sep 9, 2017 at 0:19
• 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? Sep 9, 2017 at 0:32
• @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 $-e/2$. Sep 9, 2017 at 0:44
• You're right. Thanks :) @ParamanandSingh Sep 9, 2017 at 0:49

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 $-e/2$.
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.