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UPSC NDA I and II 2020 Math Solved Paper

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Question : 105 of 120

Marks: +1, -0

What is

\underset{x→0}{lim}\frac{sinxlog(1-x)}{x^{2}}

equal to?

$-1$
Zero
-e
$-\frac{1}{e}$

Solution:

Concept:

Use standard limit results:

\underset{x \to 0}{\lim} \frac{\sin x}{x} = 1

and

\underset{x \to 0}{\lim} \frac{\log(1+x)}{x} = 1

, and apply product rule of limits.

Explanation:

Rewrite the given limit as a product:

\underset{x \to 0}{\lim} \frac{\sin x \log(1-x)}{x^2} = \underset{x \to 0}{\lim} \frac{\sin x}{x} \times \underset{x \to 0}{\lim} \frac{\log(1-x)}{x}

.

The first factor equals

1

.

For the second factor, write

1-x = 1+(-x)

, so

\underset{x \to 0}{\lim} \frac{\log(1-x)}{x} = \underset{x \to 0}{\lim} \frac{\log(1+(-x))}{x}

.

Multiply numerator and denominator by

-1

to get

\underset{x \to 0}{\lim} \frac{\log(1+(-x))}{-(-x)} = -1 \cdot \underset{x \to 0}{\lim} \frac{\log(1+(-x))}{(-x)}

.

Now

(-x) \to 0

, so by the standard limit

\underset{t \to 0}{\lim} \frac{\log(1+t)}{t} = 1

, the limit equals

-1 \times 1 = -1

.

Thus the entire product is

1 \times (-1) = -1

.

Answer:

-1

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