Question

Show by means of an example that \(\mathop {\lim }\limits_{x \to a} \,\,\left( {f\left( x \right)g\left( x \right)} \right)\) may exist even though neither \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) nor \(\mathop {\lim }\limits_{x \to a} g\left( x \right)\) exists.

Short answer

It is proved that \(\mathop {\lim }\limits_{x \to 0} \,\,f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to 0} \,\,g\left( x \right)\) does not exist, but \(\mathop {\lim }\limits_{x \to 0} \,\,\left( {f\left( x \right)g\left( x \right)} \right)\) has finite value.

Step-by-step solution

Assume the functions

Assume that \(f\left( x \right)\) be an Heaviside function, \(f\left( x \right) = H\left( x \right)\) and \(g\left( x \right)\) be defined as \(g\left( x \right) = 1 - H\left( x \right)\). It is known that the Heaviside function does not exist for \(x = 0\). So, \(\mathop {\lim }\limits_{x \to 0} \,\,f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to 0} \,\,g\left( x \right)\) does not exist.

Simplify for \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right)g\left( x \right)} \right)\)

As per the definition of\(f\left( x \right)\)and\(g\left( x \right)\), either of both the functions is 0 for any random value of\(x\). Thus, their product must be 0 at that value of\(x\).

\(\begin{aligned}\mathop {\lim }\limits_{x \to 0} \,\,\left( {f\left( x \right)g\left( x \right)} \right) &= \mathop {\lim }\limits_{x \to 0} \,\,0\\ &= 0\end{aligned}\).

Draw a conclusion

Thus, it can be concluded that \(\mathop {\lim }\limits_{x \to 0} \,\,f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to 0} \,\,g\left( x \right)\) does not exist, but \(\mathop {\lim }\limits_{x \to 0} \,\,\left( {f\left( x \right)g\left( x \right)} \right)\) has finite value.