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Question Number 63922 by raj last updated on 11/Jul/19

$$\mathrm{if}f\left(x\right)=\{\begin{array}{ll}\frac{\log (1+2ax)-\log (1-bx)}{x}& x\ne 0\\ k& x=0\end{array}$$$$\mathrm{is}\mathrm{continuous}\mathrm{at}x=0\mathrm{then}k=?$$

Commented by kaivan.ahmadi last updated on 11/Jul/19

$${lim}_{x\to 0}f\left(x\right)\stackrel{hop}{=}{lim}_{x\to 0}\frac{2a}{ln10\times (1+2ax)}-\frac{-b}{ln10\times (1-bx)}=$$$$\frac{2a}{ln10}+\frac{b}{ln10}=\frac{2a+b}{ln10}$$$$k=f\left(0\right)={lim}_{x\to 0}f\left(x\right)=\frac{2a+b}{ln10}$$

Commented by mathmax by abdo last updated on 12/Jul/19

$$atV\left(0\right)log(1+2ax)\sim \frac{2ax}{ln\left(10\right)}andlog(1-bx)\sim \frac{-bx}{ln\left(10\right)}\Rightarrow$$$$\frac{log(1+2ax)-log(1-bx)}{x}\sim \frac{2ax+bx}{xln\left(10\right)}=\frac{2a+b}{ln\left(10\right)}$$$$fcontinueat{x}_0=0\Rightarrow {lim}_{x\to 0}f\left(x\right)=f\left(0\right)=k\Rightarrow k=\frac{2a+b}{ln\left(10\right)}$$

Commented by raj last updated on 11/Jul/19

$$thankyou$$

Commented by Mikael last updated on 11/Jul/19

$$f\left(0\right)=k$$$$\underset{x\to 0}{lim}\frac{log(1+2ax)-log(1-bx)}{x}=\underset{x\to 0}{lim}\frac{log\left(\frac{1+2ax}{1-bx}\right)}{x}=\underset{x\to 0}{lim}log{\left(\frac{1+2ax}{1-bx}\right)}^{\frac{1}{x}}$$$$log\underset{u\to \mathrm{\infty }}{lim}{\left(\frac{1+2a\frac{1}{u}}{1-b\frac{1}{u}}\right)}^u=log{e}^{\underset{u\to \mathrm{\infty }}{lim}(\frac{1+2a\frac{1}{u}}{1-b\frac{1}{u}}-1).u}=log{e}^{\underset{u\to \mathrm{\infty }}{lim}\left(\frac{2a+b}{1}\right)}$$$$log{e}^{2a+b}=(2a+b).loge=\frac{(2a+b)}{ln10}$$$$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)$$$$k=\frac{2a+b}{ln10}$$

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