Question

Prove that if \(\lim _{x \rightarrow c} f(x)=0\) and \(|g(x)| \leq M\) for a fixed number \(M\) and all \(x \neq c,\) then \(\lim _{x \rightarrow c} f(x) g(x)=0\).

Short answer

According to the Squeeze theorem, the limit of the function \(f(x)g(x)\) as \(x\) approaches \(c\) is 0, given the conditions that the function \(f(x)\) approaches 0 as \(x\) approaches \(c\) and the function \(g(x)\) is bounded by a fixed number \(M\).

Step-by-step solution

Apply the Boundedness Property

Boundedness property states that if the function \(f(x)\) is converging to 0 as \(x\) approaches \(c\), then \(-M \leq f(x)g(x) \leq M\), where \(M\) is the maximum value that function \(g(x)\) can take. For the given condition \(|g(x)| \leq M\), we multiply by \(f(x)\), which is converging to 0. Thus, for any \(x \neq c\) we get \(-M.f(x) \leq f(x)g(x) \leq M.f(x)\). Because \(\lim _{x \rightarrow c} f(x)=0\), our inequalities become \(-M.0 \leq f(x)g(x) \leq M.0.\)

Apply the Squeeze Theorem

The Squeeze theorem can be used when you have three functions, \(f(x)\), \(g(x)\), and \(h(x)\) and \(f(x)\) is less than or equal to \(g(x)\) which is less than or equal to \(h(x)\) for all \(x\). If the limit as \(x\) approaches \(c\) for \(f(x)\) and \(h(x)\) is the same, then the limit as \(x\) approaches \(c\) for \(g(x)\) is also the same. In our case, we have that \(-M.0\) is less than or equal to \(f(x)g(x)\) which is less than or equal to \(M.0\). So, according to the squeeze theorem, \(\lim_{x \rightarrow c} f(x)g(x) = 0\).