Questions: A graphing calculator is recommended. Use the Squeeze Theorem to show that the limit as x approaches 0 of x^2 cos(12 pi x) = 0. Illustrate by graphing the functions f(x)=-x^2, g(x)=x^2 cos(12 pi x), and h(x)=x^2 on the same screen. Let f(x)=-x^2, g(x)=x^2 cos(12 pi x), and h(x)=x^2. Then ? ≤ cos(12 pi x) ≤ ? => ? ? ≤ x^2 cos(12 pi x) ≤ ? . Since the limit as x approaches 0 of f(x) = the limit as x approaches 0 of h(x) = 0 , by the Squeeze Theorem we have the limit as x approaches 0 of g(x) = 0

Solution Steps

To use the Squeeze Theorem to show that \(\lim_{x \rightarrow 0} x^{2} \cos (12 \pi x) = 0\), we need to find two functions \(f(x)\) and \(h(x)\) such that \(f(x) \leq g(x) \leq h(x)\) for all \(x\) in some interval around 0, and \(\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} h(x) = 0\). We can use \(f(x) = -x^2\) and \(h(x) = x^2\) because \(-1 \leq \cos(12 \pi x) \leq 1\). Therefore, \(-x^2 \leq x^2 \cos(12 \pi x) \leq x^2\). Since both \(\lim_{x \rightarrow 0} -x^2 = 0\) and \(\lim_{x \rightarrow 0} x^2 = 0\), by the Squeeze Theorem, \(\lim_{x \rightarrow 0} x^2 \cos(12 \pi x) = 0\).

To illustrate this, we can graph the functions \(f(x) = -x^2\), \(g(x) = x^2 \cos(12 \pi x)\), and \(h(x) = x^2\) on the same screen.

We define the functions \( f(x) = -x^2 \), \( g(x) = x^2 \cos(12 \pi x) \), and \( h(x) = x^2 \).

Since \(-1 \leq \cos(12 \pi x) \leq 1\), we have: \[ -x^2 \leq x^2 \cos(12 \pi x) \leq x^2 \]

We know that: \[ \lim_{x \to 0} -x^2 = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0 \] By the Squeeze Theorem, since \( -x^2 \leq x^2 \cos(12 \pi x) \leq x^2 \) and both bounds approach 0 as \( x \) approaches 0, we conclude: \[ \lim_{x \to 0} x^2 \cos(12 \pi x) = 0 \]

To illustrate this, we graph the functions \( f(x) = -x^2 \), \( g(x) = x^2 \cos(12 \pi x) \), and \( h(x) = x^2 \) on the same screen. The graph shows that \( g(x) \) is squeezed between \( f(x) \) and \( h(x) \) as \( x \) approaches 0.

Final Answer