Question

Show that $f\left(x\right)=\{\begin{array}{ccc}\frac{\cos ax-\cos bx}{x^2}& if& x\ne 0\\ \frac{1}{2}(b^2-a^2)& if& x=0\end{array}$ where a and b are real constants, is continuous at $0$.

Answer 1

$RHL=LHL=\underset{x\to 0}{lim}\frac{\cos ax-\cos bx}{x^2}$

Using L's Hospital, we get

$\underset{x\to 0}{lim}\frac{-a\sin ax+b\sin bx}{2x}$

Again Using L's Hospital, we get

$\underset{x\to 0}{lim}\frac{-a^2\cos ax+b^2\cos bx}{2}$

$\frac{b^2-a^2}{2}$

Hence it is continuous at $x=0$ as $RHL=LHL=$ value of f at $x=0$