A question about $\lim_{n\to\infty} (1+\frac{1}{n})^n = e$ with 1 replaced. $\lim_{n\to\infty} (1+\frac{1}{n})^n = e$, $\lim_{n\to\infty} (x+\frac{1}{n})^n=0,x\in(0,1)$
How come that the second limit goes to 0? If it goes to 0, shouldn't the first one go to 1?
I am thinking that for the second one to go to 0 is that 1/n goes to 0 then x^n goes to zero, and for the first the same but 1^n goes to 1 when n goes to inf.
 A: If $x<y<1$ then for large enough $n$,$$x+\frac{1}{n}<y$$ so the limit is $$<y^n\to 0$$
A: I think for $x\in(0,1)$ it goes to $0$ because there is $\epsilon>0$ and $0<q<1$, for which for all $n>\epsilon$ $x+\frac{1}{n}<q$ and $q^n$ goes to $0$.
We can take $q=\frac{x+1}{2}$ for example.
A: We have
$$\lim_{n\to\infty}(x+\frac 1n)^n = \lim_{n\to\infty}x^n(1+\frac {1/x}n)^n = \lim_{n\to\infty}x^n \lim_{n\to\infty}(1+\frac{1/x}{n})^n =  e^{1/x}\lim_{n\to\infty}x^n$$
and thus for $x>1$ the limit is $+\infty$, for $0<x<1$ the limit is $0$ and for $x=1$ the limit is $e$, as expected.
A: If $x<1$ then you can find some $N$ such that $x+\frac{1}{N} <1$. 
Set $y =x+\frac{1}{N}$.
Then, for all $n >N$ you have 
$$x^n < (n+\frac{1}{n})^n <y^n$$
and, as $0<x<y<1$ 
$$\lim x^n=\lim y^n=0$$
Therefore
$$\lim_n (x+\frac{1}{n})^n=0$$
by squeeze theorem.
A: Another way, same idea as to prove that the first one tends to $e$
$$
\left(x+\frac{1}{n}\right)^{n}=x^n\left(1+\frac{1}{nx}\right)^n=x^ne^{n\ln\left(1+ 1/nx\right)}
$$
Then
$$
\left(x+\frac{1}{n}\right)^{n}\underset{n \rightarrow +\infty}{\sim}x^n e^{1/x}
$$
$x$ is fixed and $x \in \left]0,1\right[$, hence $\ x^n \underset{n \rightarrow +\infty}{\rightarrow}0$ so
$$
\left(x+\frac{1}{n}\right)^{n}\underset{n \rightarrow +\infty}{\rightarrow}0
$$
A: Rene Schipperus has also answered why the limit is $0$ if $x \in (0,1)$. In particular the reason is not that $\frac{1}{n}$ approaches $0$, so $(x+\frac{1}{n})^n$ has the same limit as $x^n$. Instead, the reason is that we can eventually bound $(x+\frac{1}{n})^n$ away from $1$ by some $y^n$, which we know goes to $0$.
For all $0\leq x<1$, $\lim_{n\to\infty}x^n = 0$; for all $x>1$, $\lim_{n\to\infty}x^n = \infty$. In the case of $(1 + \frac{1}{n})^n$, if we could bound it below away from $1$ by some $y^n$ with $y>1$, we could conclude the limit is $\infty$. But we cannot do that. We can bound below by $1$, but then all we can say is that the limit is somewhere between $1$ and $\infty$. So the bounding method used for $x \in (0,1)$ does not extend to the $x=1$ case.
A: Let $n>\frac{2}{1-x}$. Then $x+\frac{1}{n}\leq \frac{x+1}{2}<1.$ So for $n$ large, $$(x+1/n)^n\leq \left(\frac{x+1}{2}\right)^n$$
But since $0<\frac{x+1}{2}<1$, we have $\left(\frac{x+1}{2}\right)^n\to 0$.
Alternatively, if you know that $\left(1+\frac{a}{n}\right)^n\to e^a$ for any $a$, then:
$$\left(x+\frac{1}{n}\right)^n = x^n\left(1+\frac{1/x}n\right)^n$$
and $x^n\to 0$ and $\left(1+\frac{1/x}n\right)^n\to e^{1/x}.$ so the product converges to zero.
This second approach gives you a nice upper bound for the terms, since $(1+a/n)^n$ is increasing as $n$ increases, when $a$ is positive, so $(x+1/n)^n\leq x^ne^{1/x}.$
A: Take the logarithm of your sequence and compute its limit:
$$
\lim_{n\to\infty}n\log\left(x+\frac{1}{n}\right)=
\lim_{t\to0^+}\frac{\log(x+t)}{t}
$$
provided the latter limit exists. Now $\log(x+t)=\log x+\log(1+t/x)$ and the further substitution $u=t/x$ yields
$$
\lim_{u\to0^+}\frac{\log x+\log(1+u)}{ux}=-\infty
$$
because the numerator has limit $\log x<0$.
By continuity of the exponential,
$$
\lim_{n\to\infty}\left(x+\frac{1}{n}\right)^{\!n}=0
$$
The same argument tells us that, for $x>1$, the limit is $\infty$.
