Question

36: Prove that \(\mathop {\lim }\limits_{x \to 2} \frac{1}{x} = \frac{1}{2}\).

Short answer

It is proved that \(\mathop {\lim }\limits_{x \to 2} \left( {\frac{1}{x}} \right) = \frac{1}{2}\).

Step-by-step solution

Guess a value for \(\delta \)

Let \(\varepsilon > 0\)be a given positive number.

We need to determine a number \(\delta > 0\) such that when\(\left| {\frac{1}{x} - \frac{1}{2}} \right| < \varepsilon \) whenever \(0 < \left| {x - 2} \right| < \delta \). So, it is represented as:

\(\begin{array}{c}\left| {\frac{1}{x} - \frac{1}{2}} \right| = \left| {\frac{{2 - x}}{{2x}}} \right|\\ = \frac{{\left| {x - 2} \right|}}{{\left| {2x} \right|}} < \varepsilon \end{array}\)

We want to determine the positive number \(C\) such that;

\(\begin{array}{c}\frac{1}{{\left| {2x} \right|}} < C\\\frac{{\left| {x - 2} \right|}}{{\left| {2x} \right|}} < C\left| {x - 2} \right|\end{array}\)

It is possible to make \(C\left| {x - 2} \right| < \varepsilon \) by taking \(\left| {x - 2} \right| < \frac{\varepsilon }{C} = \delta \).

We can restrict \(x\) to lie in the interval such that;

\(\begin{array}{c}\left| {x - 2} \right| < 1\\ - 1 < x - 2 < 1\\1 < x < 3\end{array}\)

Therefore,

\(\begin{array}{c}1 > \frac{1}{x} > \frac{1}{3}\\\frac{1}{6} < \frac{1}{{2x}} < \frac{1}{2}\\\frac{1}{{\left| {2x} \right|}} < \frac{1}{2}\end{array}\)

Hence, \(C = \frac{1}{2}\) is a suitable choice for the constant.

Therefore, we must choose \(\delta = \min \left\{ {1,2\varepsilon } \right\}\).

Prove that \(\mathop {\lim }\limits_{x \to 2} \frac{1}{x} = \frac{1}{2}\)

Let \(\varepsilon > 0\)be a given positive number. Consider \(\delta = \min \left\{ {1,2\varepsilon } \right\}\). When \(0 < \left| {x - 2} \right| < \delta \) then;

\(\begin{array}{l}\left| {x - 2} \right| < 1\\1 < x < 3\\\frac{1}{{\left| {2x} \right|}} < \frac{1}{2}{\rm{ }}\left\{ {{\mathop{\rm from}\nolimits} \,\,{\mathop{\rm part}\nolimits} \left( a \right)} \right\}\end{array}\)

Moreover, \(\left| {x - 2} \right| < 2\varepsilon \) therefore;

\(\begin{array}{c}\left| {\frac{1}{x} - \frac{1}{2}} \right| = \frac{{\left| {x - 2} \right|}}{{\left| {2x} \right|}}\\ < \frac{1}{2} \cdot 2\varepsilon \\ = \varepsilon \end{array}\)

This proves that \(\mathop {\lim }\limits_{x \to 2} \left( {\frac{1}{x}} \right) = \frac{1}{2}\).

Thus, it is proved that \(\mathop {\lim }\limits_{x \to 2} \left( {\frac{1}{x}} \right) = \frac{1}{2}\).