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UPSC NDA I and II 2020 Math Solved Paper

Question : 108 of 120

Marks: +1, -0

What is

\underset{x→0}{lim}\frac{3^{x}+3^{-x}-2}{x}

equal to ?

Solution:

Concept:

Use the standard limit

\underset{x \to 0}{\lim} \frac{a^{x} - 1}{x} = \ln a

and split the given expression into two such forms.

Explanation:

Rewrite the numerator:

3^{x} + 3^{-x} - 2 = (3^{x} - 1) + (3^{-x} - 1)

Then the limit becomes

\underset{x \to 0}{\lim} \frac{(3^{x} - 1) + (3^{-x} - 1)}{x}

Using sum rule:

\underset{x \to 0}{\lim} \frac{3^{x} - 1}{x} + \underset{x \to 0}{\lim} \frac{3^{-x} - 1}{x}

Note that

3^{-x} = (3^{-1})^{x}

. So the second limit is

\underset{x \to 0}{\lim} \frac{(3^{-1})^{x} - 1}{x} = \ln(3^{-1})

-\ln 3

The first limit is

\ln 3

Sum:

\ln 3 + (-\ln 3) = 0

Answer:

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