Question

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

Answer 1

Since f(x) is continuous  at x = 0 

${\displaystyle \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)}$

Now ${\displaystyle \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{e^{2x}-1}{ax}}$

= ${\displaystyle \frac{2}{a}\left[\underset{x\to 0}{lim}\frac{e^{2x}-1}{2x}\right]}$

= ${\displaystyle \frac{2}{a}\left[\log e\right]}$   [∵ ${\displaystyle \underset{x\to 0}{lim}\frac{a^x-1}{x}=\log a}$]

= ${\displaystyle \frac{2}{a}\left[1\right]=\frac{2}{a}}$

${\displaystyle \underset{x\to 0^{-}}{lim}f\left(x\right)=\frac{2}{a}}$

Now ${\displaystyle \underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{\log (1+7x)}{bx}}$

= ${\displaystyle \frac{7}{b}\left[\underset{x\to 0}{lim}\frac{\log (1+7x)}{7x}\right]}$

= ${\displaystyle \frac{7}{b}\left[1\right] [\because \underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1]}$

∴ ${\displaystyle \underset{x\to 0^{+}}{lim}f\left(x\right)=\frac{7}{b}}$

Also f(0) = -1

∴ ${\displaystyle \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)}$

∴ ${\displaystyle \frac{2}{a}=\frac{7}{b}=1}$

∴ a = 2 , b = 7