Question

If $f\left(x\right)=\{\begin{array}{c}\frac{\sin ax}{\sin bx},x\ne 0\\ \frac{a}{b},x=0\end{array}$
Test the continuity of function at $x=0$.

Answer 1

$f\left(x\right)=\frac{\sin ax}{\sin bx}$

$f\left(0\right)=\frac{a}{b}.......\left(1\right)$ (Given)

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\frac{\sin ax}{\sin bx}$

$=\underset{h\to 0}{lim}\frac{\sin a(0+h)}{\sin b(0+h)}=\underset{h\to 0}{lim}\frac{\sin ah}{\sin bh}$

$=\underset{h\to 0}{lim}(\frac{\sin ah}{ah}\times ah\times \frac{1}{\frac{\sin bh}{bh}\times bh})=\frac{a}{b}\times \underset{h\to 0}{lim}\left(\frac{\sin ah}{ah}\right)\times \frac{1}{\underset{h\to 0}{lim}\left(\frac{\sin bh}{bh}\right)}$

$=\frac{a}{b}\times 1\times \frac{1}{1}$

$=\frac{a}{b}.....\left(2\right)$

again

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\frac{\sin ax}{\sin bx}$

$=\underset{h\to 0}{lim}\frac{\sin a(0-h)}{\sin b(0-h)}=\underset{h\to 0}{lim}\frac{-\sin ah}{-\sin bh}$

$=\underset{h\to 0}{lim}\frac{\sin ah}{\sin bh}$

$=\underset{h\to 0}{lim}(\frac{\sin ah}{ah}\times ah\times \frac{1}{\frac{\sin bh}{bh}\times bh})=\frac{a}{b}\times \underset{h\to 0}{lim}\left(\frac{\sin ah}{ah}\right)\times \frac{1}{\underset{h\to 0}{lim}\left(\frac{\sin bh}{bh}\right)}$

$=\frac{a}{b}\times 1\times \frac{1}{1}=\frac{a}{b}.....\left(3\right)$

From equation (1),(2),(3)

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

$\therefore$ Function $f\left(x\right)$ is continuous at $x=0$

Hence proved