Q:

Limits : solving indeterminate form

Hey can someone help in working out the right hand limit.  The curly brackets indicate fractional part
$$\lim_{n\to0^+}\bigg(\frac {(1+\left\{n\right\})^{\frac {1}{\left\{n\right\}}}}{e}\bigg)^{\frac {1}{\left\{n\right\}}}$$
I am having a confusion that the quantity inside the bracket is tending to 1 but whatever it be, it will always be less than 1 ( I analysed and found that the derivative at x =0+ is negative) , thus anything less than 1raised to the power infinity is 0 , but the answer is coming different , pls enlighten and clarify my doubt

A:

I don't think the "fractional function" matters, especially as we are approaching zero, where everything is only a "fraction"
$$\lim _{n \to 0} \left(\frac{(1+n)^{1/n}}{e}\right)^{1/n} = L \approx6.066...\text{ according to Wolfram Alpha}$$
$$\lim _{n \to 0} \left(\frac{1}{n}*\ln\frac{(1+n)^{1/n}}{e}\right) = \ln L$$
$$\lim _{n \to 0} \left(\frac{1}{n}* \left( \ln\frac{(1+n)^{1/n}}{e}\right)\right) = \lim _{n \to 0} \left(\frac{1}{n}*(\ln(1+n)^{1/n} - \ln(e))\right)$$
$$\lim_{n \to 0} \frac{1}{n}*\ln\left((1+n)^{1/n}\right) - \frac{1}{n}*1 = \lim_{n \to 0} \frac{1}{n^2}*\ln(1+n) - \frac{1}{n}$$
Using the MacLaurin series for $ln(1+n) = n - n^2/2 + n^3/3 -n^4/4 ...$
$$\lim_{n \to 0} \frac{1}{n^2}*(n - n^2/2 + n^3/3 - n^4/4 ...) - \frac{1}{n} = \lim_{n \to 0} 1/n - 1/2 + n/3 - n^2/4 + n^3/5 .. - 1/n$$
And at $0$ all of the terms with $n$ will become $0$ and $1/n$ cancels leaving us with 
$$-1/2 = \ln L $$
$$ L = e^{-1/2} = 0.60653065$$

