Let f(x)= (1+ x a )- (1- x b ) x ,x 0 Find the value of f at x = 0. So that f becomes continuous at x = 0.

Given, f(x)= (1+ x a )- (1- x b ) x ,x 0 If f(x) is continuous at x = 0, then _ x 0 f(x)=f(0) _ x 0 (1+ x a )- (1- x b ) x =f(0) _ x 0 (1+ x a ) ax a - (1- x b ) bx b =f(0) _ x 0 (1+ x a ) x a - (-1 b ) _ x 0 (1- x b ) -x b =f(0) 1 a 1- (-1 b ) 1=f(0) [Using : _ x 0 log(1+x) x =1 ] 1 a +1 b =f(0) a+b ab =f(0)

Let f(x)= (1+ x a )- (1- x b ) x ,x 0 Find the value of f at x = …

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Let $\text{f}\text{(x)}=\frac{\log\Big(1+\frac{\text{x}}{\text{a}}\Big)-\log\Big(1-\frac{\text{x}}{\text{b}}\Big)}{\text{x}},\text{x}\neq0$ Find the value of f at x = 0. So that f becomes continuous at x = 0.

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Given, $\text{f}\text{(x)}=\frac{\log\Big(1+\frac{\text{x}}{\text{a}}\Big)-\log\Big(1-\frac{\text{x}}{\text{b}}\Big)}{\text{x}},\text{x}\neq0$
If f(x) is continuous at x = 0, then
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{\log\Big(1+\frac{\text{x}}{\text{a}}\Big)-\log\Big(1-\frac{\text{x}}{\text{b}}\Big)}{\text{x}}\end{pmatrix}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{\log\Big(1+\frac{\text{x}}{\text{a}}\Big)}{\frac{\text{ax}}{\text{a}}}-\frac{\log\Big(1-\frac{\text{x}}{\text{b}}\Big)}{\frac{\text{bx}}{\text{b}}}\end{pmatrix}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{\log\Big(1+\frac{\text{x}}{\text{a}}\Big)}{\frac{\text{x}}{\text{a}}}\end{pmatrix}-\Big(-\frac{1}{\text{b}}\Big)\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{\log\Big(1-\frac{\text{x}}{\text{b}}\Big)}{\frac{-\text{x}}{\text{b}}}\end{pmatrix}=\text{f}(0)$
$\Rightarrow\frac{1}{\text{a}}\times1-\Big(-\frac{1}{\text{b}}\Big)\times1=\text{f}(0)$ $\Big[\text{Using :}\lim_{\text{x} \rightarrow 0}\frac{\text{log(1}+\text{x)}}{\text{x}}=1\Big]$
$\Rightarrow\frac{1}{\text{a}}+\frac{1}{\text{b}}=\text{f}(0)$
$\Rightarrow\frac{\text{a}+\text{b}}{\text{ab}}=\text{f}(0)$

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