Question

38: If H is the Heaviside function defined in section 2.2, prove, using Definition 2, that \(\mathop {\lim }\limits_{t \to 0} H\left( t \right)\) does not exist.

(Hint: Use an indirect proof as follows. Suppose that the limit is L. Take \(\varepsilon = \frac{1}{2}\) in the definition of a limit and try to arrive at a contradiction.)

Short answer

It is proved that \(\mathop {\lim }\limits_{t \to 0} H\left( t \right)\) does not exist.

Step-by-step solution

Precise Definition of a limit

Consider \(f\) as afunctiondefined on some open interval that contains the number \(a\), except possibly at \(a\) itself.

Then, we say that the limit of \(f\left( x \right)\) as \(x\)approaches \(a\) is \(L\), and it can be written as\(\mathop {\lim }\limits_{x \to a} f\left( x \right) = L\).

If for all numbers \(\varepsilon > 0\) there exists a number \(\delta > 0\) such that if \(0 < \left| {x - a} \right| < \delta \) then \(\left| {f\left( x \right) - L} \right| < \varepsilon \).

Prove that \(\mathop {\lim }\limits_{t \to 0} H\left( t \right)\) does not exist

Considerthat \(\mathop {\lim }\limits_{t \to 0} H\left( t \right) = L\). Let \(\varepsilon = \frac{1}{2}\) be a given positive number there is a number \(\delta > 0\) such that;

\(\begin{array}{c}0 < \left| t \right| < \delta \\\left| {H\left( t \right) - L} \right| < \frac{1}{2}\\L - \frac{1}{2} < H\left( t \right) < L + \frac{1}{2}\end{array}\)

For \(0 < t < \delta \), \(H\left( t \right) = 1\), therefore;

\(\begin{array}{c}1 < L + \frac{1}{2}\\L > \frac{1}{2}\end{array}\)

For \( - \delta < t < 0\), \(H\left( t \right) = 0\), therefore;

\(\begin{array}{c}L - \frac{1}{2} < 0\\L < \frac{1}{2}\end{array}\)

Which contradicts that \(L > \frac{1}{2}\).

Hence, \(\mathop {\lim }\limits_{t \to 0} H\left( t \right)\) does not exist.

Thus, it is proved that \(\mathop {\lim }\limits_{t \to 0} H\left( t \right)\) does not exist.