Question

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

Answer 1

${lim}_{x\to 0}\frac{e^{2x}-1}{ax}={lim}_{x\to 0}\frac{e^{2x}-1}{2ax}\times 2=\frac{2}{a}$

${lim}_{x\to 0}\frac{\log (1+7x)}{b}={lim}_{x\to 0}\frac{\log (1+7x)}{7bx}\times 7=\frac{7}{b}$

Given that $f\left(x\right)$ is continuous at $x=0$

$\Rightarrow \frac{2}{a}=1$ or $a=2$

$\Rightarrow \frac{7}{b}=1$ or $b=7$

$\therefore a=2,b=7$