$\lim_{x\to 0^+}[1+[x]]^{2/x}$,where [x] denotes greatest integer function less than or equal to x Find the value of $$\lim\limits_{x\to 0^+}[1+[x]]^{\frac2x}$$where $[x]$ denotes greatest integer function less than or equal to $x$. 



My attempt: 
I calculated $[1+[x]]$ to be $1$ as $x\to 0^+$.  Now I am stuck.  Please help me.
For $0&lt;x\ll 1$, $[1+[x]] = 1$. Thus $[1+[x]]^{2/x} = 1^{2/x} = 1$.This gives 
$$ \lim_{x→ 0^+}[1+[x]]^{2/x} = \lim_{x→ 0^+} 1 = 1$$
Substitute $y=\frac1x$, so as $x\to0^{+}, y\to\infty$. 
So we get $$\lim\limits_{y\to\infty}[1+[\frac{1}{y}]]^{2y}=\lim\limits_{y\to\infty}([1+[\frac1y]]^{y})^2=1^{2.\infty}=1$$