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Let \[e\] denote the base of the natural logarithm. Then find the value of the real number \[a\} for which the right-hand limit \[\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{{\left( {1 - x} \right)}^{\dfrac{1}{x}}} - {e^{ - 1}}}}{{{x^a}}}\] is equal to a nonzero real number.

Answer
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Hint: We will apply the right-hand limit \[\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{{\left( {1 - x} \right)}^{\dfrac{1}{x}}} - {e^{ - 1}}}}{{{x^a}}}\]. Then we will apply the formula \[{e^{ln\left( x \right)}} = x\] in place \[{\left( {1 - x} \right)^{\dfrac{1}{x}}}\]. Then we will expand the natural logarithm and simplify it. After that apply the formula \[{e^{a + b}} = {e^a} \cdot {e^b}\] and expand \[{e^{\left( { - \dfrac{x}{2} - \dfrac{{{x^2}}}{3} - \cdots } \right)}}\]. We know that, the denominator of a fraction can never zero. For any real value of \[a\] except \[1\], the value of \[\mathop {\lim }\limits_{x \to {0^ + }} {x^{a - 1}}\] will be zero. So the possible value of \[a\] will be zero.
Formula Used:
\[{e^{ln\left( x \right)}} = x\]
\[\ln {\left( x \right)^m} = m\,\ln \left( x \right)\]
\[ln\left( {1 - x} \right) = - x - \dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} - ......\]
Complete step by step solution:
The given right-hand limit is, \[\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{{\left( {1 - x} \right)}^{\dfrac{1}{x}}} - {e^{ - 1}}}}{{{x^a}}}\].
Let’s simplify the given limit.
Let \[L\] be the value of the limit.
 \[L = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{{\left( {1 - x} \right)}^{\dfrac{1}{x}}} - {e^{ - 1}}}}{{{x^a}}}\]
Now apply the exponent and logarithmic rule .
\[L = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{e^{\ln \left( {{{\left( {1 - x} \right)}^{\dfrac{1}{x}}}} \right)}} - {e^{ - 1}}}}{{{x^a}}}\]
Apply the logarithmic property .
\[L = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{e^{\dfrac{1}{x}\ln \left( {1 - x} \right)}} - {e^{ - 1}}}}{{{x^a}}}\]
Expand the natural \[\log \] term as a series.
\[L = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{e^{\dfrac{1}{x}\left( { - x - \dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} - \cdots } \right)}} - \dfrac{1}{e}}}{{{x^a}}}\]
Simplify the above equation.
\[L = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{e^{\left( { - 1 - \dfrac{x}{2} - \dfrac{{{x^2}}}{3} - \cdots } \right)}} - \dfrac{1}{e}}}{{{x^a}}}\]
\[L = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{e^{ - 1}} \cdot {e^{\left( { - \dfrac{x}{2} - \dfrac{{{x^2}}}{3} - \cdots } \right)}} - \dfrac{1}{e}}}{{{x^a}}}\]
Apply the formula \[{e^x} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \cdots \]
\[ \Rightarrow L = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{e^{ - 1}}\left( {1 + \left( { - \dfrac{x}{2} - \dfrac{{{x^2}}}{3}} \right) + \dfrac{{{{\left( { - \dfrac{x}{2} - \dfrac{{{x^2}}}{3}} \right)}^2}}}{{2!}} + \cdots } \right) - \dfrac{1}{e}}}{{{x^a}}}\]
Factoring out the common factor.
\[ \Rightarrow L = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{e^{ - 1}}\left( {x\left( { - \dfrac{1}{2} - \dfrac{x}{3}} \right) + \dfrac{{{x^2}{{\left( { - \dfrac{1}{2} - \dfrac{x}{3}} \right)}^2}}}{{2!}} + \cdots - 1} \right)}}{{{x^a}}}\]
Divide numerator and denominator by \[x\]
\[ \Rightarrow L = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{e^{ - 1}}\left( {\left( { - \dfrac{1}{2} - \dfrac{x}{3}} \right) + \dfrac{{x{{\left( { - \dfrac{1}{2} - \dfrac{x}{3}} \right)}^2}}}{{2!}} + \cdots - 1} \right)}}{{{x^{a - 1}}}}\]
The limit exists if \[{x^{a - 1}} = 1\].
So,
\[{x^{a - 1}} = {x^0}\]
Compare the power of \[x\].
\[a - 1 = 0\]
\[ \Rightarrow \]\(a = 1\]
Hence, the value of the real number \[a\] is 1
Note: The right-hand limit of a function is as the variable approaches to a negligible slight increment.
Students are often get confused with the expansion of .The correct expansion is \[ln\left( {1 - x} \right) = - x - \dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} - ......\].