16. - / 5.88 points determine whether ( f^{prime}(0) ) exists. f(x)=left{\begin{array}{ll} x sin \frac{9}{x}…

Explanation:

Step1: Recall the definition of the derivative

The derivative of a function (f(x)) at (x = 0) is given by (f^{\prime}(0)=\lim_{x\rightarrow0}\frac{f(x)-f(0)}{x - 0}). Since (f(0) = 0) and (f(x)=x\sin\frac{9}{x}) for (x\neq0), then (\frac{f(x)-f(0)}{x - 0}=\frac{x\sin\frac{9}{x}-0}{x}=\sin\frac{9}{x}) for (x\neq0).

Step2: Evaluate the limit

We need to find (\lim_{x\rightarrow0}\sin\frac{9}{x}). Let (t=\frac{9}{x}). As (x\rightarrow0), (t\rightarrow\pm\infty). We know that the function (y = \sin t) oscillates between (- 1) and (1) as (t\rightarrow\pm\infty). (\lim_{x\rightarrow0}\sin\frac{9}{x}) does not exist because (\sin\frac{9}{x}) oscillates infinitely as (x\rightarrow0).

Answer:

(f^{\prime}(0)) does not exist.