One method is via the substitution $x = 1/y$. As $x \to \infty$, $y \to 0+$.
The given function can be written as
$$
\begin{eqnarray*}
\frac 1 y \left[ (1+y)^{1/y} - {\mathrm e} \right]
&=&
\frac{1}{y} \left[ \exp\left(\frac{\ln (1+y)}{y} \right) - \mathrm e \right]
\\&=&
\frac{\mathrm e}{y} \left[ \exp\left(\frac{\ln (1+y) - y}{y} \right) - 1 \right]
\\ &=&
\mathrm e \cdot \frac{\ln(1+y)-y}{y^2}
\cdot \frac{ \exp\left(\frac{\ln (1+y) - y}{y} \right) - 1 }{\frac{\ln (1+y) - y}{y}}
\end{eqnarray*}
$$
Can you go from here? The last factor approaches $1$ by the standard limit $\frac{{\mathrm e}^z -1}{z} \to 1$ as $z \to 0$. The limit of the middle factor can be evaluated by L'Hôpital's rule.
There are multiple mistakes on both sides.
I said $0$ multiplied by anything should equal to $0$ even when $x$ approaches infinity.
This is not true! Indeed, $\lim_{x \to \infty} \frac 1x = 0$, but $\lim_{x \to \infty} \frac 1x \cdot x^2 = \infty$. Therefore, you were wrong on this front, but it turns out your answer was correct, and you acknowledged it by writing $\left[\frac 6x\right] = 0$ as $x$ approaches infinity. This is a stronger condition than the limit existing and equalling zero.
$\left[\frac{6}{x}\right] = \frac 6x - k$ for some $0 \leq k < 1$.
Note that we know what $k$ is, because $\frac 6x < 1$ for $x > 6$, so in fact $\left[\frac 6x \right] = 0$ for $x > 6$ , which simply makes $k = \frac 6x$! This allows us to work out what happens when $x \to \infty$, so you can conclude the answer here, and it is not $-\infty$, but in fact $0$.
Then my friend said $0 \times \infty$ can't be zero.
Once again not true, take $\frac 1{x^2}$ and the sequence $x$ whose product goes to zero although one goes to infinity.
For the answers:
Zero times infinity is not zero. However, the first sequence $\left[\frac 6x\right ]\frac x3$ is zero after $x>6$ because the first fraction is $0$, so the limit is $0$. Note that terms being equal to zero after some time, is stronger than the limit being zero. This absorbs the sequence $\frac x3$ regardless of what properties it may have.
The second you know.
For the third, we have to be more careful : we have an infinity i.e. $\left[\frac 6x\right]$ and a zero i.e. $\frac x3$ convergent sequence. Now, here $\infty \times 0$ confusion comes in, which is sorted by setting appropriate bounds on $\left[\frac 6x\right]$.
What bounds? Obviously, $\frac 6x -1 \leq \left[\frac 6x\right] \leq \frac 6x $.
Setting these in, we get :
$$
\left(\frac 6x -1 \right)\frac x3\leq \left[\frac{6}{x}\right]\frac x3 \leq \frac 6x\frac{x}{3}
$$
Now all we are left to notice is that the left and right hand side have limit $2$ as $x$ converges to $0$, hence the middle also has the same limit by squeeze theorem.
Best Answer
The order of the two limits is important. If you reverse the order you get a different answer. You need to do the inner one first then the outer one.
As the inner limit doesn't depend on $n$ you can rewrite it as: $$\lim_{n\rightarrow\infty}(2^{-n}\lim_{k\rightarrow\infty}\frac{1}{2^{-n}+2^{-k}})$$ The inner limit is then just $\frac{1}{2^{-n}}$ which would cancel giving you $$\lim_{n\rightarrow\infty}1=1$$
Or did you mean a limit where both are happening at the same time? If so you shouldn't use two different variables.