Convergence of the sequence $n ((ea)^{1/n} - a^{1/n})$ for positive $a$ Suppose $a$ is positive. Consider the sequence 
$$a_n=n\big((ea)^{\frac {1}{n}}- a^{\frac {1}{n}}\big) , n\geq 1$$
Then
A. $\displaystyle\lim_{n \to \infty} a_n$ does not exist
B. $\displaystyle\lim_{n \to \infty} a_n = e$
C. $\displaystyle\lim_{n \to \infty} a_n = 0$
D. None of above
On writing out few terms and taking arbitrary $a$, it seems answer is C, but I am not sure
Thanks
 A: Consider the limit
$$\lim_{n \to \infty} n((ea)^{1 \over n} - a ^{1 \over n}) $$
$$ = \lim_{n \to \infty} n((e^{1 \over n}a^{1 \over n} - a ^{1 \over n}) $$$$ = \lim_{n \to \infty} n(a^{1 \over n}(e^{1 \over n} - 1)) $$
$$ =\lim_{n \to \infty} \frac{a^{1 \over n}(e^{1 \over n} - 1)}{1 \over n} $$
Substitute $x = {1 \over n}$
$$ =\lim_{x \to 0} \frac{a^{x}(e^{x} - 1)}{x} $$
Using L'Hospital's rule for $0 \over0 $ we get
$$ =\lim_{x \to 0} a^x(\ln(a)*(e^{x} - 1) +e^x) = 1 $$
A: $\begin{array}\\
a_n
&=n\big((ea)^{\frac {1}{n}}- a^{\frac {1}{n}}\big)\\
&=na^{\frac {1}{n}}\big(e^{\frac {1}{n}}- 1\big)\\
&=ne^{\frac {\ln a}{n}}\big(e^{\frac {1}{n}}- 1\big)\\
&\approx n(1+\frac {\ln a}{n}+O(1/n^2))\big(1+\frac {1}{n}+O(1/n^2)- 1\big)\\
&= n(1+\frac {\ln a}{n}+O(1/n^2))\big(\frac {1}{n}+O(1/n^2)\big)\\
&= (1+\frac {\ln a}{n}+O(1/n^2))\big(1+O(1/n)\big)\\
&=1+O(1/n)\\
&\to 1\\
\end{array}
$
A: Notice that $\displaystyle\lim_{n \to \infty}a_n$ $= \displaystyle\lim_{n \to \infty}\dfrac{(ea)^{1/n}-a^{1/n}}{\tfrac{1}{n}}$ $= \displaystyle\lim_{x \to 0}\dfrac{(ea)^{x}-a^x}{x}$ 
$= \displaystyle\lim_{x \to 0}\left[\dfrac{(ea)^{x}-1}{x} - \dfrac{a^{x}-1}{x}\right]$ $= \displaystyle\lim_{x \to 0}\dfrac{(ea)^{x}-1}{x} - \lim_{x \to 0}\dfrac{a^{x}-1}{x}$, 
provided that both of these limits exist. Can you compute these limits? 
Hint: Think about the definition of the derivative.
