in the following exercises, use the squeeze theorem to prove the limit. 223. \\( \\lim _{x \\rightarrow 0}…

in the following exercises, use the squeeze theorem to prove the limit. 223. \\( \\lim _{x \\rightarrow 0} x^{2} \\cos (2 \\pi x)=0 \\) 224. \\( \\lim _{x \\rightarrow 0} x^{3} \\sin \\left(\\frac{\\pi}{x}\\right)=0 \\)

in the following exercises, use the squeeze theorem to prove the limit. 223. \\( \\lim _{x \\rightarrow 0} x^{2} \\cos (2 \\pi x)=0 \\) 224. \\( \\lim _{x \\rightarrow 0} x^{3} \\sin \\left(\\frac{\\pi}{x}\\right)=0 \\)

Answer

Explanation:

Step1: Analyze the range of the trigonometric function

For (223), we know that (- 1\leqslant\cos(2\pi x)\leqslant1). Multiply each part of the inequality by (x^{2}) (since (x^{2}\geqslant0) for all real (x)), we get (-x^{2}\leqslant x^{2}\cos(2\pi x)\leqslant x^{2}). For (224), we know that (-1\leqslant\sin(\frac{\pi}{x})\leqslant1). Multiply each part of the inequality by (x^{3}) (when (x) is near (0), and consider the sign of (x). If (x>0), (x^{3}>0); if (x < 0), (x^{3}<0)). We get (-x^{3}\leqslant x^{3}\sin(\frac{\pi}{x})\leqslant x^{3})

Step2: Apply the Squeeze Theorem

For (223), (\lim_{x\rightarrow0}(-x^{2})=0) and (\lim_{x\rightarrow0}(x^{2}) = 0). By the Squeeze Theorem, (\lim_{x\rightarrow0}x^{2}\cos(2\pi x)=0) For (224), (\lim_{x\rightarrow0}(-x^{3})=0) and (\lim_{x\rightarrow0}(x^{3})=0). By the Squeeze Theorem, (\lim_{x\rightarrow0}x^{3}\sin(\frac{\pi}{x})=0)

Answer:

For (223), (\lim_{x\rightarrow0}x^{2}\cos(2\pi x)=0) is proved. For (224), (\lim_{x\rightarrow0}x^{3}\sin(\frac{\pi}{x})=0) is proved.