Limits that lead to the indeterminate forms ${{1}^{\infty }},{{0}^{0}},{{\infty }^{0}}$ can sometimes be solved taking logarithm first and then using L’Hospital’s rule. Then, $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,{{x}^{\dfrac{1}{(1-x)}}}$ is equal to

Hint: For solving this problem, we will first convert the limit into desirable form using logarithmic function. After converting it to desirable form we can easily evaluate the limit using L'Hospital's rule to obtain the answer.

Complete step by step answer:
Let $\underset{x\to a}{\mathop{\lim }}\,{{\left( f(x) \right)}^{g(x)}}$ is in the form of ${{\infty }^{0}}$, it can be written as ${{e}^{\underset{x\to a}{\mathop{\lim }}\,g(x)\ln f(x)}}={{e}^{L}}$ where $L=\underset{x\to a}{\mathop{\lim }}\,\dfrac{\ln f(x)}{\dfrac{1}{g(x)}}$ is $\dfrac{\infty }{\infty }$ form and can be solved using l'hospital's rule.
As per the problem statement, we have to evaluate $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,{{x}^{\dfrac{1}{(1-x)}}}$. So, we first simplify it by letting the given part as L. Now, we can mathematically express it as:
\[L=\underset{x\to 1}{\mathop{\lim }}\,{{x}^{\dfrac{1}{x-1}}}\]
Let us assume y as x-1 to further simplify the expression. Now, when x is tending to 1 then y is tending to 0.
$\therefore L=\underset{y\to 0}{\mathop{\lim }}\,{{(1+y)}^{\dfrac{1}{y}}}\ldots (1)$
Now, as we know that for a logarithmic function $\log {{x}^{n}}=n\log x$. So, by applying the logarithmic function on equation (1), we get
\[\begin{align}
  & \ln L=\ln \left( \underset{y\to 0}{\mathop{\lim }}\,{{(1+y)}^{\dfrac{1}{y}}} \right) \\
 & \ln L=\underset{y\to 0}{\mathop{\lim }}\,\ln \left\{ {{(1+y)}^{\dfrac{1}{y}}} \right\} \\
 & \ln L=\underset{y\to 0}{\mathop{\lim }}\,\left\{ \dfrac{1}{y}\ln (1+y) \right\}\ldots (2) \\
\end{align}\]
So, now by using the expansion of $\ln (1+x)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-..........\infty $, we can expand equation (2) to get,
$\begin{align}
  & \ln L=\underset{y\to 0}{\mathop{\lim }}\,\left\{ \dfrac{1}{y}\left( y-\dfrac{{{y}^{2}}}{2}+\dfrac{{{y}^{3}}}{3}-.......\infty \right) \right\} \\
 & \ln L=\underset{y\to 0}{\mathop{\lim }}\,\left\{ 1-\dfrac{y}{2}+\dfrac{{{y}^{2}}}{3}-\dfrac{{{y}^{3}}}{4}.......\infty \right\} \\
\end{align}$
So, by putting the limit as $\underset{x\to 0}{\mathop{\lim }}\,\left( 1-\dfrac{x}{2}+\dfrac{{{x}^{2}}}{3}-\dfrac{{{x}^{3}}}{4}........\infty \right)=1$ because all the other terms would be zero except for the first term. So, finally we get, $\ln L=1$.
Now, taking the antilog of both sides, we get
$\begin{align}
  & {{e}^{\ln L}}={{e}^{1}} \\
 & \therefore L=e \\
\end{align}$
Hence, the correct limit is e.

Note: The key step for solving this problem is the formulation of the given statement into desired form by using logarithmic function. Once the student obtains the desired logarithmic form, the limit of complex function is evaluated. Students must take care of the calculations that are involved in transforming the equation.