Find $\lim_{x \to \infty} (\frac{1}{e} - \frac{x}{x+1})^{x}$ Find $\lim_{x \to \infty} (\frac{1}{e} - \frac{x}{x+1})^{x}$
This is of the form $(\frac{1}{e} - \frac{1}{(1/∞) + 1})^{∞}$
Now I'm stuck..
How should I go further.
If I find the limit of the function inside the bracket, it is $(1/e - 1)$, which is a negative number less than $1$. It is approx $-0.632$. So its power $(=∞)$ should make the number $0$.
But it is given in answer that the limit doesn't exist. 
I think I'm missing some basic concept.
Any suggestion$?$
 A: Note that $$\frac1e-\frac{x}{x+1} = \frac1e-\frac{x+1-1}{x+1} = \frac1e-1+\frac{1}{x+1}$$
$$= \underbrace{-\left(1-\frac1e \right)}_{\textrm{negative constant}} + \underbrace{\frac{1}{x+1}}_{\textrm{tends to zero}}$$
so the expression $\left(\frac1e-\frac{x}{x+1}\right)^x$ is undefined for some arbitrarily large values of $x$ (for example, if $x=n+\frac12$ for large integers $n$).
This means the limit can't exist. The reason is that if the limit did exist, then it would would also equal $$\lim\limits_{n\to\infty}\left(\frac1e-\frac{a_n}{a_n+1}\right)^{a_n}$$
for sequence $(a_n)$ which increases without bound. But we have demonstrated such a sequence for which every one of the terms is undefined.

Addendum: I point out that notation is critical here. In writing "$x\to\infty$", the implication is that $x$ grows without bound taking on all real values in the process.

However, if you used the notation "$n\to\infty$" instead (and used the variable $n$ instead of $x$), convention is that you have a sequence and $n$ grows without bound taking on integer values only.
This is purely conventional usage and is not strictly well-defined, but it is certainly largely accepted and used. Context would clarify, and in this case it is critical.

Addendum 2: If you allow complex values of the expression and multiple-valuedness of them, then the limit does indeed exist and equals zero. Here's why. We may consider only $x$ sufficiently large so that the base is always negative (as described above). Then we can write
$$\left(\frac1e-\frac{x}{x+1}\right)^x = \left(r(x)e^{\pi i}\right)^x$$
where $r(x) = 1-\frac1e - \frac1{x+1}$ is strictly positive. This expression is multiple-valued and takes values (for integral $k$)
$$r(x)^x\cdot e^{(2k+1)\pi xi} = r(x)^x\cdot e^{i\theta_k(x)}$$ where $\theta_k(x)=(2k+1)\pi x$ for integral $k$.
As others have shown, $r(x)^x\to 0$, so the value of $\theta_k(x)$ doesn't matter; the angles grow without bound, but the magnitude shrinks to zero, so in the end the value spirals to $0$ as $x\to\infty$.
This is true for all of the multiple values (the values corresponding to fixed $k$), so the limit exists independent of the choice of $k$. The expressions are well-defined (multiple-valuedly) for all $x$, and all tend to zero. So the limit is zero.
A: If your limit converges to $L$ then
$$
\ln L
 = \ln \left(\lim_{x \to \infty} \left(\frac{1}{e} - \frac{x}{x+1}\right)^x \right)
 = \lim_{x \to \infty}  \ln \left(\left(\frac{1}{e} - \frac{x}{x+1}\right)^x \right)
$$
and now exponent $x$ can be pulled outside the log...
UPDATE
Note that $1/e < 1$  so for $x$ large enough the denominator is negative. Hence, if $x \in \mathbb{R}$, the limit does not exist. If, on the other hand, $x \in \mathbb{N}$, this is well-defined and the calculations will make sense...
A: Assuming $f(x):\mathbb R\to \mathbb R$ the limit exists and it is equal to zero indeed according to the definition of limit (refer for example to "Rudin's Principles of Mathematical Analysis")
$$\Big(\lim_{x\rightarrow \infty} f(x) = L \Big)\iff \Big(\forall \varepsilon >0\, \exists \delta: \forall x\in D\quad x>\delta \implies \vert f(x)-L\vert <\varepsilon \Big)$$
where $D$ is the domain where $f(x)$ can be evaluated, that is
$$D = \mathbb{R} \setminus \{x:\Im(f(x))\neq0\}$$
therefore since
$$\frac{1}{e} - \frac{x}{x+1} \to \frac 1e-1=\frac{1-e}{e}, \quad a=\left|\frac{1-e}{e}\right|<1 $$
then
$$\left|\frac{1}{e} - \frac{x}{x+1}\right|^{x} \to a^\infty=0$$
and therefore according to the given definition of limit, since $D$ is a not empty subset of $\mathbb R$ with $\infty$ as limit point, we have that

$$\left(\frac{1}{e} - \frac{x}{x+1}\right)^{x} \to 0$$

To a similar discussion with many good answer from many users, refer to the related


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*What is $\lim_{x \to 0}\frac{\sin(\frac 1x)}{\sin (\frac 1 x)}$ ? Does it exist?
