Find the limit of $\lim_{n\to \infty} (1+n)^{\frac1n}$ Find the following limit:
$$\lim_{n\to \infty} (1+n)^{\frac1n}$$
I know that 
$\lim_{n\to 0} (1+n)^{\frac1n}= e$.  
 A: We have $(1+n)^{1/n}>1$ for all $n$, hence let $(1+n)^{1/n} = 1+a_n$,where $a_n > 0$. Further, we have
$$1+n = (1+a_n)^n \geq 1+ \dfrac{n(n-1)}2a_n^2 \implies a_n^2 \leq \dfrac2{n-1}$$
This means $$\lim_{n \to \infty} a_n^2 \leq \lim_{n \to \infty} \dfrac2{n-1} = 0 \implies \lim_{ n \to \infty} a_n = 0$$ Hence, we obtain that
$$\lim_{n \to \infty} (1+n)^{1/n} = \lim_{n \to \infty} \left(1 + a_n\right) = 1+ \lim_{n \to \infty} a_n = 1$$
A: $$\lim_{n\to\infty}(1+n)^{\frac{1}{n}}=$$
$$\lim_{n\to\infty}\exp\left(\ln\left((1+n)^{\frac{1}{n}}\right)\right)=$$
$$\lim_{n\to\infty}\exp\left(\frac{1}{n}\ln\left(1+n\right)\right)=$$
$$\lim_{n\to\infty}\exp\left(\frac{\ln\left(1+n\right)}{n}\right)=$$
$$\lim_{n\to\infty}\exp\left(\frac{\frac{\text{d}}{\text{d}x}\left(\ln\left(1+n\right)\right)}{\frac{\text{d}}{\text{d}x}\left(n\right)}\right)=$$
$$\lim_{n\to\infty}\exp\left(\frac{\frac{1}{1+n}}{1}\right)=$$
$$\lim_{n\to\infty}\exp\left(\frac{1}{1+n}\right)=$$
$$\exp\left(\lim_{n\to\infty}\frac{1}{1+n}\right)=\exp(0)=e^0=1$$
A: $$\lim_\limits{n\to\infty} (1+n)^\frac{1}{n}$$
$$=\lim_\limits{n\to\infty} e^{\ln(1+n)^\frac{1}{n}}$$
$$=\lim_\limits{n\to\infty} e^{\frac{\ln(1+n)}{n}}$$
$$=e^{\lim_\limits{n\to\infty} \frac{\ln(1+n)}{n}}$$
Now $\lim_\limits{n\to\infty} \frac{\ln(1+n)}{n}$ is indeterminate since it is of the form $\frac{\infty}{\infty}$.
Hence by L'hospital's rule,
$$\lim_\limits{n\to\infty} \frac{\ln(1+n)}{n}=\lim_\limits{n\to\infty} \frac{\frac{1}{1+n}}{1}=0$$
Required limit $=e^0=1$
A: Let $n=\frac{1}{t}\implies t\to 0$ as $n\to \infty$, hence 
$$\lim_{n\to \infty}(1+n)^{\frac 1n}=\lim_{t\to 0}\left(1+\frac 1t\right)^{t}$$
$$=\lim_{t\to 0}\left(\frac{1+t}{t}\right)^{t}$$
$$=\lim_{t\to 0}\frac{(1+t)^t}{t^t}$$
$$=\frac{\color{blue}{\lim_{t\to 0}(1+t)^t}}{\color{red}{\lim_{t\to 0}(t^t)}}=\frac{1}{1}=\color{red}{1}$$

Proofs: It is obvious $\color{blue}{\lim_{t\to 0}(1+t)^t=1^0=1}$, & 
  $$\color{red}{\lim_{t\to 0}(t^t)}=\exp\lim_{t\to 0}t\ln (t)=\exp\lim_{t\to 0}\frac{\frac{d}{dt}(\ln (t))}{\frac{d}{dt}(1/t)}=\exp\lim_{t\to 0}\frac{1/t}{-1/t^2}=\exp\lim_{t\to 0}(-t)=e^0=1$$

A: You start with 
$\lim\limits_{n\rightarrow \infty} (1+n)^{\frac{1}{n}}$.
It should be the same as the following:
$e^{\ln(\lim\limits_{n\rightarrow \infty} (1+n)^{\frac{1}{n}})}$
$=e^{\frac{1}{n}\ln(\lim\limits_{n\rightarrow \infty} (1+n))}$
Can you figure out where to go from here?
