Question

If $f\left(x\right)=\frac{e^{2x}-1}{ax}$, for $x<0,a\ne 0$
$=1$, for $x=0$
$=\frac{\log (1+7x)}{bx}$, for $x>0,b\ne 0$
Is continuous at $x=0$ then find $a$ and $b$.

Answer 1

Consider the given function.
$f\left(x\right)=\frac{e^{2x}-1}{ax}$, for $x<0,a\ne 0$
$f\left(x\right)=1$, for $x=0$
$f\left(x\right)=\frac{log(1+7x)}{bx}$, for $x>0,b\ne 0$
Since, the function is continuous at $x=0$
So, $R.H.L=L.H.L=f\left(0\right)$
Therefore,
${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{-}}\frac{e^{2x}-1}{ax}$
${lim}_{x\to 0^{-}}[\frac{2}{a}\times \frac{e^{2x}-1}{2x}]=1$
$\frac{2}{a}\times 1=1$
$a=2$
Similarly,
${lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}\frac{log(1+7x)}{bx}$
${lim}_{x\to 0^{+}}\frac{7}{b}\frac{log(1+7x)}{7x}=1$
$\frac{7}{b}=1$
$b=7$
Hence, this is the answer.