Question

Show that $f\left(x\right)=\{\begin{array}{cc}\frac{\cos ax-\cos bx}{x^2},& \text{if}x\ne 0\\ \frac{1}{2}(b^2-a^2),& \text{if}x=0\end{array}$

where $a$ and $b$ are real constants, is continuous at $x=0$.

Answer 1

$R.H.L={lim}_{x\to 0^{+}}f\left(x\right)={lim}_{\to 0^{+}}\frac{\cos ax-\cos bx}{x^2}$

$={lim}_{\to 0^{+}}\frac{2\sin \left(\frac{a+b}{2}\right)x\sin \left(\frac{b-a}{2}\right)x}{x^2}$

$={lim}_{\to 0^{+}}\frac{2\sin \left(\frac{a+b}{2}\right)x\sin \left(\frac{b-a}{2}\right)x}{x.x}$

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

$=\frac{1}{2}(b^2-a^2)$

$L.H.L={lim}_{x\to 0^{-}}f\left(x\right)={lim}_{\to 0^{-}}\frac{\cos ax-\cos bx}{x^2}$

$={lim}_{\to 0^{-}}\frac{2\sin \left(\frac{a+b}{2}\right)x\sin \left(\frac{b-a}{2}\right)x}{x^2}$

$={lim}_{\to 0^{-}}\frac{2\sin \left(\frac{a+b}{2}\right)x\sin \left(\frac{b-a}{2}\right)x}{x.x}$

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

$=\frac{1}{2}(b^2-a^2)$

$R.H.L=L.H.L=f\left(x\right)$

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