Question

If f(x) defined by f(x) = {log(1+ax)-log(1-bx)}/x when x not=0 f(x) k if x=0 Find the value of k

Answer 1

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given a function f(x) continuous at x = 0

To Find:

• We have to find the value of k

Solution:

We have been given a Function f(x) defined

$f(x) = \begin{cases} \dfrac{log(1+ax) - log(1-bx) }{x} & \text{ if } \: x \neq 0 \\ \\</p><p>\: \: \: k & \text{ if } \: x = 0 \end{cases}$

is continuous at x = 0

_______________________________

Since f(x) is continuos at x = 0

∴ $f(0) = \lim_{x \to 0}{f(x)}$

$\implies k = \lim_{x \to 0} {f(x)}$

Putting value of f(x) when $x \neq 0$

$\implies k = \lim_{x \to 0} \left ( \dfrac{log(1+ax)-log(1-bx)}{x} \right )$

Breaking denominator x in separate fraction

$\implies k = \lim_{x \to 0} \left ( \dfrac{log(1+ax)}{x} - \dfrac{log(1-bx)}{x} \right )$

On separating limits

$\implies k = \underbrace{ \lim_{x \to 0} \left ( \dfrac{log(1+ax)}{x} \right ) }_{A}- \underbrace{\lim_{x \to 0} \left ( \dfrac{log(1-bx)}{x} \right ) }_{B}$

$\implies k = A - B$ __________ ( 1 )

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Finding the value of part A

$A = \lim_{x \to 0} \left ( \dfrac{log(1+ax)}{x} \right )$

Multiplying and Dividing by 'a' on RHS

$A =a \lim_{x \to 0} \left ( \dfrac{log(1+ax)}{ax} \right )$

$A = a$ ____________( 2 )

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Finding the value of part B

$B = \lim_{x \to 0} \left ( \dfrac{log(1-bx)}{x} \right )$

Multiplying and Dividing by ( -b ) on RHS

$B =(-b) \lim_{x \to 0} \left ( \dfrac{log(1+(-bx))}{(-bx)} \right )$

$B = (-b)$ ___________ ( 3 )

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Putting values of A and B from equation ( 2 ) and ( 3 ) in equation ( 1 )

$\implies k = A + B$

$\implies k = a - ( -b )$

$\implies k = a + b$

Hence value of k is ( a + b )

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$\huge\mathfrak\red{Identity \: \: Used :}$

$\fbox{\lim_{x \to a}[f(x) +g(x)] = \lim_{x \to a}f(x) + \lim_{x \to a} g(x) }$

$\fbox{\lim_{x \to 0} \left ( \dfrac{log(1+x)}{x} \right ) = 1}$

Answer 2

Step-by-step explanation:

hope it will help you.............