Answer
2.718283
Work Step by Step
The original function for this problem is $$f(x)=(1+x)^\frac{1}{x}$$
So, the hole, or undefined point, of the function as $x$ is approaching $0$, which can be $x\rightarrow 0$, which notes that the point will be labeled as $$0,y$$
So, when it comes to the limitation of this function when $x=0$, it can be expressed as the following:
$$\begin{matrix}
_{x\rightarrow 0}^{\lim}\textrm{f(x)}&=_{x\rightarrow 0}^{\lim}\mathrm{(1+x)^\frac{1}{x}}\\
&=e\\
&\approx 2.71828
\end{matrix}$$
First, we will find the points to the left of $0$, which the first 6 $x$-values and $f(x)$'s are the following:
$$\begin{matrix}
x & &f(x) \\
-.1 & (1+(-.1))^\frac{1}{(-.1)} &2.867972 \\
-.01 & (1+(-.01))^\frac{1}{(-.01)} &2.731999 \\
-.001 & (1+(-.001))^\frac{1}{(-.001)} &2.719642 \\
-.0001 & (1+(-.0001))^\frac{1}{(-.0001)} &2.718418 \\
-.00001 & (1+(-.00001))^\frac{1}{(-.00001)} &2.718295 \\
-.000001 & (1+(-.000001))^\frac{1}{(-.000001)} &2.718283
\end{matrix}$$
Second, we will find the points to the left of $0$, which the first 6 $x$-values and $f(x)$'s are the following:
$$\begin{matrix}
x & &f(x) \\
.1 & (1+(.1))^\frac{1}{(.1)} &2.593742\\
.01 & (1+(.01))^\frac{1}{(.01)} &2.704814\\
.001 & (1+(.001))^\frac{1}{(.001)} &2.716942\\
.0001 & (1+(.0001))^\frac{1}{(.0001)} &2.718146\\
.00001 & (1+(.00001))^\frac{1}{(.00001)} &2.718268\\
.000001 & (1+(.000001))^\frac{1}{(.000001)} &2.718280
\end{matrix}$$
Therefore, the official hole, or undefined point is at $$(0,e)$$
