Question

Let $e$ denote the base of the natural logarithm. The value of the real number $a$ for which the right-hand limit $\underset{x\to 0^{+}}{\lim }\frac{{\left(1-x\right)}^{{\displaystyle \frac{1}{x}}}-e^{-1}}{x^a}$is equal to a nonzero real number, is _____

Answer 1

Using properties of logarithm, transform the given limit into standard form

$\underset{x\to 0^{+}}{\lim }\frac{{\left(1-x\right)}^{{\displaystyle \frac{1}{x}}}-e^{-1}}{x^a}$$=$$\underset{x\to 0^{+}}{\lim }\frac{{\left(1-x\right)}^{{\displaystyle \frac{1}{x}\ln \left(e\right)}}-e^{-1}}{x^a}$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \ln \left(e\right)=1\right]$

$\Rightarrow$ $=\underset{x\to 0^{+}}{\lim }\frac{{\left(e\right)}^{{\displaystyle \frac{1}{x}\ln \left(1-x\right)}}-e^{-1}}{x^a}$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because a^{\ln \left(b\right)}=b^{\ln \left(a\right)}\right]$

$\Rightarrow$ $=e^{-1}\underset{x\to 0^{+}}{\lim }\frac{e^{\left({\displaystyle \frac{\ln \left(1-x\right)}{x}}+1\right)}-1}{x^a}$ $...\left(i\right)$

Multiply and divide by $\frac{\ln \left(1-x\right)}{x}+1$ inside the limit to bring into standard form

$\Rightarrow$$\underset{x\to 0^{+}}{\lim }\frac{{\left(1-x\right)}^{{\displaystyle \frac{1}{x}}}-e^{-1}}{x^a}$ $=\frac{1}{e}\underset{x\to 0^{+}}{\lim }\frac{e^{\left(\frac{\ln \left(1-x\right)}{x}+1\right)}-1}{{\displaystyle \frac{\ln \left(1-x\right)}{x}}+1}\times \frac{\frac{\ln \left(1-x\right)}{x}+1}{x^a}$

Let $\frac{\ln \left(1-x\right)}{x}+1=t$

$\Rightarrow$$\underset{x\to 0^{+}}{\lim }\frac{{\left(1-x\right)}^{{\displaystyle \frac{1}{x}}}-e^{-1}}{x^a}$$=\frac{1}{e}\underset{t\to 0^{+}}{\lim }\frac{e^t-1}{t}\times \underset{x\to 0^{+}}{\lim }\frac{{\displaystyle \frac{\ln \left(1-x\right)}{x}}+1}{x^a}$

$\Rightarrow$ $=\frac{1}{e}\underset{x\to 0^{+}}{\lim }\frac{{\displaystyle \frac{\ln (1-x)}{x}}+1}{x^a}$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \underset{x\to 0}{\lim }\frac{e^x-1}{x}=1\right]$

$\Rightarrow$ $=\frac{1}{e}\underset{x\to 0^{+}}{\lim }\frac{{\displaystyle \ln (1-x)}+x}{x^{a+1}}$ $...\left(ii\right)$

Apply L'Hospital's rule to find the limit

$\Rightarrow$$\underset{x\to 0^{+}}{\lim }\frac{{\left(1-x\right)}^{{\displaystyle \frac{1}{x}}}-e^{-1}}{x^a}$$=\frac{1}{e}\underset{x\to 0^{+}}{\lim }\frac{{\displaystyle \frac{d}{dx}}\left(\ln (1-x\right)+x)}{{\displaystyle \frac{d}{dx}}{x}^{a+1}}$

$\Rightarrow$ $=\frac{1}{e}\underset{x\to 0^{+}}{\lim }\frac{{\displaystyle \frac{-1}{1-x}}+1}{\left(a+1\right)x^a}$

$\Rightarrow$ $=\frac{1}{e}\underset{x\to 0^{+}}{\lim }\frac{-1}{\left(x-1\right)\left(a+1\right)}\times \frac{x}{x^a}$

$\Rightarrow$ $=\frac{-1}{e}\underset{x\to 0^{+}}{\lim }\frac{x^{1-a}}{\left(-1\right)\left(a+1\right)}$

$\Rightarrow$ $=\frac{1}{e\left(a+1\right)}\underset{x\to 0^{+}}{\lim }{x}^{1-a}$

For positive value of $a$ limit tends to $0$

For negative value of $a$limit tends to $\infty$

As the limit is said to be non zero real number the value of $1-a$ must be zero.

$\Rightarrow$$1-a=0$

$\Rightarrow$ $a=1$

Hence, the value of $a$ is $1$ for the given right hand limit to be a non zero real number.