What is the value of the limit \[\mathop {lim}\limits_{x \to 0} {\left( {1 - ax} \right)^{\dfrac{1}{x}}}\] ?
A. \[{e^{ - a}}\]
B. \[e\]
C. \[{e^a}\]
D. 1

Hint: First, simplify the given expression of the limit using the standard formula of the limit \[\mathop {lim}\limits_{x \to a} {\left[ {f\left( x \right)} \right]^{g\left( x \right)}} = {e^{\mathop {lim}\limits_{x \to a} g\left( x \right)\left[ {f\left( x \right) - 1} \right]}}\] . Then solve the expression of the limit to reach the required answer.

Formula Used:
\[\mathop {lim}\limits_{x \to a} {\left[ {f\left( x \right)} \right]^{g\left( x \right)}} = {e^{\mathop {lim}\limits_{x \to a} g\left( x \right)\left[ {f\left( x \right) - 1} \right]}}\]

Complete step by step solution:
The given expression of the limit is \[\mathop {lim}\limits_{x \to 0} {\left( {1 - ax} \right)^{\dfrac{1}{x}}}\].
Let's consider \[L\] to be the value of the above expression.
Then,
\[L = \mathop {lim}\limits_{x \to 0} {\left( {1 - ax} \right)^{\dfrac{1}{x}}}\]
Now apply the standard formula of limit \[\mathop {lim}\limits_{x \to a} {\left[ {f\left( x \right)} \right]^{g\left( x \right)}} = {e^{\mathop {lim}\limits_{x \to a} g\left( x \right)\left[ {f\left( x \right) - 1} \right]}}\].
We get,
\[L = {e^{\mathop {lim}\limits_{x \to 0} \dfrac{1}{x}\left[ {\left( {1 - ax} \right) - 1} \right]}}\]
Simplify the above equation.
\[L = {e^{\mathop {lim}\limits_{x \to 0} \dfrac{1}{x}\left[ { - ax} \right]}}\]
\[ \Rightarrow L = {e^{\mathop {lim}\limits_{x \to 0} \left[ { - a} \right]}}\]
Now apply the limit.
\[L = {e^{ - a}}\]
Therefore, the value of the given limit is
\[\mathop {lim}\limits_{x \to 0} {\left( {1 - ax} \right)^{\dfrac{1}{x}}} = {e^{ - a}}\]

Hence the correct option is A.

Note: Students often make mistakes while solving the limit. They apply the limit without cancelation of the common factors. Because of that, the value of the limit became 0, infinite or indeterminant.