ₓarrow 0}(1-ax)¹/x} ?

Correct Answer is : e−ae^{-a}e−a

Correct Answer is : e−ae^{-a}e−a

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PUNE MCA Exam 2015 Question Paper with answers for online practice

Question : 15 of 59

Marks: +1, -0

\lim\limits_{x\rightarrow 0}(1-ax)^{\frac{1}{x}}

Solution:

Concept:

The standard limit

\lim\limits_{x\to 0} (1 + kx)^{1/x} = e^k

Explanation:

Rewrite the expression:

(1 - ax)^{1/x} = [1 + (-a)x]^{1/x}

x \to 0

, the limit takes the form

\lim\limits_{x\to 0} [1 + (-a)x]^{1/x} = e^{-a}

Thus, the required limit is

e^{-a}

Answer:

e^{-a}

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