Question

What is the $\displaystyle \lim_{x \to 1+}{{x}^{\left( \dfrac{1}{1-x} \right)}}$ as x approaches 1 from the right side?

Answer 1

Hint: Assume the given limit as L. Now, take natural log, i.e. log to the base e both the sides, take the log function inside the limit and use the property of log given as $\ln \left( {{a}^{m}} \right)=m\ln \left( a \right)$ to simplify. Now, apply the L Hospital’s rule to simplify the limit $\dfrac{\ln x}{1-x}$. Use the formula $\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}$ and remove the log function from both the sides to get the answer.

Complete step by step solution:
Here we have been provided with the limit expression $\displaystyle \lim_{x \to 1+}{{x}^{\left( \dfrac{1}{1-x} \right)}}$ and we are asked to find its limit value as x is tending to 1 from rightward direction. Let us assume this limit as L, so we have,
$\Rightarrow L=\displaystyle \lim_{x \to 1+}{{x}^{\left( \dfrac{1}{1-x} \right)}}$
Now, if we will try to substitute the value of x = 1 in the given expression then the limit will become of the form ${{1}^{\infty }}$. To evaluate such limits we have to follow the following steps.
Taking the natural log function, i.e. log to the base e on both the sides we get,
$\Rightarrow \ln L=\ln \left( \displaystyle \lim_{x \to 1+}{{x}^{\left( \dfrac{1}{1-x} \right)}} \right)$
We can take the log function inside the limit expression so we get,
$\Rightarrow \ln L=\displaystyle \lim_{x \to 1+}\left[ \ln \left( {{x}^{\left( \dfrac{1}{1-x} \right)}} \right) \right]$
Using the property of log given as $\ln \left( {{a}^{m}} \right)=m\ln \left( a \right)$ we get,
$\begin{align}
  & \Rightarrow \ln L=\displaystyle \lim_{x \to 1+}\left[ \dfrac{1}{1-x}\times \ln x \right] \\
 & \Rightarrow \ln L=\displaystyle \lim_{x \to 1+}\left[ \dfrac{\ln x}{1-x} \right] \\
\end{align}$
Now, we can see that in the R.H.S we have the limit of the form $\dfrac{0}{0}$ so we can apply the L Hospital’s rule to simplify the limit, we need to differentiate the numerator and the denominator till it comes in a determinate form, so we get,
\[\Rightarrow \ln L=\displaystyle \lim_{x \to 1+}\left[ \dfrac{\dfrac{d\left( \ln x \right)}{dx}}{\dfrac{d\left( 1-x \right)}{dx}} \right]\]
Using the formulas $\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}$, $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ and the fact that the derivative of a constant is 0 we get,
\[\begin{align}
  & \Rightarrow \ln L=\displaystyle \lim_{x \to 1+}\left[ \dfrac{\left( \dfrac{1}{x} \right)}{0-1} \right] \\
 & \Rightarrow \ln L=\displaystyle \lim_{x \to 1+}\left[ \dfrac{-1}{x} \right] \\
\end{align}\]
The above expression is now a determinate form of the limit so substituting x = 1 we get,
\[\begin{align}
  & \Rightarrow \ln L=\left[ \dfrac{-1}{1} \right] \\
 & \Rightarrow \ln L=-1 \\
\end{align}\]
We can write -1 in the R.H.S as $\ln \left( {{e}^{-1}} \right)$ so we get,
\[\Rightarrow \ln L=\ln \left( {{e}^{-1}} \right)\]
Removing the log function from both the sides and equating their argument due to the fact that the base of log is same on both the sides, we get,
\[\begin{align}
  & \Rightarrow L={{e}^{-1}} \\
 & \therefore L=\dfrac{1}{e} \\
\end{align}\]
Hence, the value of the given expression of limit is $\dfrac{1}{e}$.

Note: There is a direct formula that you may apply to solve the question in less steps. The formula says the if we have the limit expression of the form $\displaystyle \lim_{x \to a}\left[ {{\left( f\left( x \right) \right)}^{g\left( x \right)}} \right]$ which is of the form ${{1}^{\infty }}$ at x = a then we can directly write the limit as \[{{e}^{\displaystyle \lim_{x \to a}\left[ g\left( x \right)\times \left( f\left( x \right)-1 \right) \right]}}\]. This formula is derived using the same logarithmic approach that we have used to solve the question. Remember that the L Hospital’s rule is only applicable if the function $\dfrac{p\left( x \right)}{q\left( x \right)}$ is of the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$.