Question

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

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 \(223\), \(\lim_{x
ightarrow0}(-x^{2})=0\) and \(\lim_{x
ightarrow0}(x^{2}) = 0\). By the Squeeze Theorem, \(\lim_{x
ightarrow0}x^{2}\cos(2\pi x)=0\)
For \(224\), \(\lim_{x
ightarrow0}(-x^{3})=0\) and \(\lim_{x
ightarrow0}(x^{3})=0\). By the Squeeze Theorem, \(\lim_{x
ightarrow0}x^{3}\sin(\frac{\pi}{x})=0\)

Answer:

For \(223\), \(\lim_{x
ightarrow0}x^{2}\cos(2\pi x)=0\) is proved. For \(224\), \(\lim_{x
ightarrow0}x^{3}\sin(\frac{\pi}{x})=0\) is proved.