Question

1. - / 5.92 points determine whether ( f^{prime}(0) ) exists. ( f(x)=left{\begin{array}{ll}x^{2} sin \frac{7}{x} & \text { if } x

eq 0 \\ 0 & \text { if } x=0end{array}
ight. ) ( \bigcirc f^{prime}(0) ) does exist. ( \bigcirc f^{prime}(0) ) does not exist.

Explanation:

Step1: Recall the definition of the derivative

The derivative of a function \(f(x)\) at \(x = a\) is given by \(f^{\prime}(a)=\lim_{x
ightarrow a}\frac{f(x)-f(a)}{x - a}\). For \(a = 0\) and \(f(0)=0\), we have \(f^{\prime}(0)=\lim_{x
ightarrow0}\frac{f(x)-f(0)}{x-0}=\lim_{x
ightarrow0}\frac{x^{2}\sin\frac{7}{x}-0}{x}\).

Step2: Simplify the limit expression

Simplify \(\lim_{x
ightarrow0}\frac{x^{2}\sin\frac{7}{x}}{x}\). Using the rule of fraction simplification \(\frac{x^{2}\sin\frac{7}{x}}{x}=x\sin\frac{7}{x}\) for \(x
eq0\).

Step3: Use the Squeeze Theorem

We know that \(- 1\leqslant\sin\frac{7}{x}\leqslant1\) for \(x
eq0\). Multiply each part of the inequality by \(x\) (when \(x>0\), the inequality direction remains; when \(x < 0\), the inequality direction is reversed). We get \(-|x|\leqslant x\sin\frac{7}{x}\leqslant|x|\).
Since \(\lim_{x
ightarrow0}-|x| = 0\) and \(\lim_{x
ightarrow0}|x|=0\), by the Squeeze Theorem \(\lim_{x
ightarrow0}x\sin\frac{7}{x}=0\).

Answer:

\(f^{\prime}(0)\) does exist.