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, although \(\mathop {\lim }\limits_{x \to a} \,\,f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to a} \,\,g\left( x \right)\) does not exist for any integral value of \(a\), but \(\mathop {\lim }\limits_{x \to a} \,\,\left( {f\left( x \right) + g\left( x \right)} \right)\) has finite value.

Step-by-step solution

Describe the given information

Assume equal and opposite greatest integer function for\(f\left( x \right)\)and\(g\left( x \right)\), such that\(f\left( x \right) = \left(\kern-0.15em\left( x

\right)\kern-0.15em\right)\)and\(g\left( x \right) = - \left(\kern-0.15em\left( x

\right)\kern-0.15em\right)\).

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

For any value of a, the sum of\(f\left( x \right)\)and\(g\left( x \right)\)is 0, as these are equal and opposite. Thus, we get:

\(\begin{aligned}{c}\mathop {\lim }\limits_{x \to a} \,\,\left( {f\left( x \right) + g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} \,\,\left(\kern-0.15em\left( x

\right)\kern-0.15em\right) - \left(\kern-0.15em\left( x

\right)\kern-0.15em\right)\\ = 0\end{aligned}\).

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

As\(f\left( x \right)\)and\(g\left( x \right)\)are greatest integer function, then at particular value of\(a\), say 3,\(\mathop {\lim }\limits_{x \to {3^ + }} \,\,f\left( x \right) = 3\), but\(\mathop {\lim }\limits_{x \to {3^ - }} \,\,f\left( x \right) = 2\). So,\(\mathop {\lim }\limits_{x \to 3} \,\,f\left( x \right)\)does not exist.

Similarly, if\(\mathop {\lim }\limits_{x \to {3^ + }} \,\,g\left( x \right) = - 3\), but \(\mathop {\lim }\limits_{x \to {3^ - }} \,\,g\left( x \right) = - 2\). So, \(\mathop {\lim }\limits_{x \to 3} \,\,g\left( x \right)\) does not exist. Thus \(\mathop {\lim }\limits_{x \to a} \,\,f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to a} \,\,g\left( x \right)\) does not exist for any integral value of \(a\).