17. - / 5.92 points determine whether ( f^{prime}(0) ) exists. ( f(x)=left{\begin{array}{ll}x^{2} sin…

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\rightarrow a}\frac{f(x)-f(a)}{x - a}). For (a = 0) and (f(0)=0), we have (f^{\prime}(0)=\lim_{x\rightarrow0}\frac{f(x)-f(0)}{x-0}=\lim_{x\rightarrow0}\frac{x^{2}\sin\frac{7}{x}-0}{x}).

Step2: Simplify the limit expression

Simplify (\lim_{x\rightarrow0}\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\neq0).

Step3: Use the Squeeze Theorem

We know that (- 1\leqslant\sin\frac{7}{x}\leqslant1) for (x\neq0). 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\rightarrow0}-|x| = 0) and (\lim_{x\rightarrow0}|x|=0), by the Squeeze Theorem (\lim_{x\rightarrow0}x\sin\frac{7}{x}=0).

Answer:

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