Calculating limit including greatest integer function $\displaystyle\lim_{n\to\infty} \left((1.5)^n +[(1+0.0001)^{10000}]^n\right)^{1/n}$   where [] means greatest integer function
I simplified the expression  using binomial expansion as
$ \left((1.5)^n+ 2^n\right)^{1/n}$
$L = \left((1.5)^n+ 2^n\right)^{1/n}$
$\log (L)= \frac{1}{n} \log(1.5)^n +\frac{1}{n} \log(2)^n  $
$\log(L) =\log(3)$
This gives limit as $3$ but answer stated is e. I'm not able to find a flaw here
 A: One mistake is the following:
$L = \left(1.5^n+ 2^n\right)^{1/n}\implies\log (L)= \frac{1}{n} \left(\log(1.5)^n + \log(2)^n\right). $
Correct is:
$L = \left((1.5)^n+ 2^n\right)^{1/n}\implies\log (L)= \frac{1}{n} \log\left(1.5^n + 2^n\right). $
Next,
$(1+0.0001)^{10000}\approx e = 2.718\ldots\ .\ $ Therefore, $\lfloor{(1+0.0001)^{10000}\rfloor} = 2.$
Furthermore for $a>0,\ b>0$,
$$\displaystyle\lim_{n\to\infty} \left(a^n +b^n\right)^{1/n} = \max\{a,b\}.$$
Since $2 = \lfloor{(1+0.0001)^{10000}\rfloor} > 1.5,\ $ we therefore have:
$$\displaystyle\lim_{n\to\infty} \left((1.5)^n +\lfloor(1+0.0001)^{10000}\rfloor^n\right)^{1/n} = \lim_{n\to\infty} \left((1.5)^n +2^n\right)^{1/n} = 2.$$
A: $
\begin{align*}
L &= (1.5^n + 2^n)^{\frac{1}{n}} \\
&= 2 (1 + 0.75^n)^{\frac{1}{n}} \\
\Rightarrow \lim_{n \rightarrow \infty} L &= 2
\end{align*}
 $
A: All you need to see is that $s_{}=(1+\frac{1}{10^3})^{10^3}$ is greater than $\frac{3}{2}$ by expanding in Binomial series, as I wrote in the comments, as the first two terms sum to $2$, so that $\frac{3}{2s}<1$. Then, pull it out of the brackets:
$$
s(1+(\frac{3}{2s})^{n})^n = s e^{n \log (1+(\frac{3}{2s})^n)} \approx s e^{n (\frac{3}{2s})^n} \to_n s
$$
You can add more terms from Maclaurin series expansion to get upper and lower bounds, but $n \alpha^n$ converges to $0$ regardless for $|\alpha|<1$, so the results follows.
