What is {x→1}{lim}f(x)-1/g(x) equal to?

Correct Answer is : ln(ab)2ln(ab)/22ln(ab)

Correct Answer is : ln(ab)2ln(ab)/22ln(ab)

Test Index

UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 53-54

Consider the following for the next items that follow:

Let

f(x)=\frac{a^{x-1}+b^{x-1}}{2}

and g(x) = x - 1.

Question : 53 of 120

Marks: +1, -0

What is

\underset{x→1}{lim}\frac{f(x)-1}{g(x)}

equal to?

Solution:

Concept:

Use the standard limit

\lim_{y \to 0} \frac{a^y - 1}{y} = \ln a

and properties of logarithms.

Explanation:

Given

f(x) = \frac{a^{x-1} + b^{x-1}}{2}

and

g(x) = x-1

We need

L = \lim_{x \to 1} \frac{f(x)-1}{g(x)}

Substitute

f(x)

L = \lim_{x \to 1} \frac{\frac{a^{x-1}+b^{x-1}}{2} - 1}{x-1}

Let

y = x-1

, so as

x \to 1

y \to 0

. Then

L = \lim_{y \to 0} \frac{\frac{a^y + b^y}{2} - 1}{y}

Simplify:

L = \lim_{y \to 0} \frac{a^y + b^y - 2}{2y} = \frac{1}{2} \lim_{y \to 0} \frac{(a^y - 1) + (b^y - 1)}{y}

Using limit laws:

L = \frac{1}{2} \left( \lim_{y \to 0} \frac{a^y - 1}{y} + \lim_{y \to 0} \frac{b^y - 1}{y} \right)

Since

\lim_{y \to 0} \frac{a^y - 1}{y} = \ln a

and

\lim_{y \to 0} \frac{b^y - 1}{y} = \ln b

we get

L = \frac{1}{2} (\ln a + \ln b) = \frac{1}{2} \ln(ab)

Answer:

\frac{\ln(ab)}{2}

(Option B)

Go to Question:

Prev Question

Next Question