Question

Prove that if \(\lim _{x \rightarrow c} f(x)=0,\) then \(\lim _{x \rightarrow c}|f(x)|=0\).

Short answer

The limit of the absolute value of the given function \(f(x)\) as \(x\) approaches \(c\) is shown to be zero. The proof is based on the limit definition and the properties of absolute value.

Step-by-step solution

Understanding the limit definition

Firstly, start from the definition of limit, i.e., for any \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever \(0 < |x - c| < \delta\), then \(|f(x) - 0| < \epsilon\). This definition stems from the main point being proved that \(\lim _{x \rightarrow c} f(x)=0.\)

Using the definition of absolute value

Recall the definition of an absolute value as \(|a| = a\) if \(a >= 0\) and \(-a\) if \(a < 0\). This property allows us to state that \(|f(x)|<= |f(x) - 0|<\epsilon \) whenever \(0< |x-c|<\delta \). The left inequality is always true because absolute value is always greater or equal to the actual value and the right inequality comes from the previous step.

Completing the proof

For the final step, note that we have shown for any \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever \(0 < |x - c| < \delta\), \(|f(x)| < \epsilon\). This is the definition of \(\lim _{x \rightarrow c}|f(x)|=0\). So, the proof is completed.