Question

1. - / 5.88 points determine whether ( f^{prime}(0) ) exists. f(x)=left{\begin{array}{ll} x sin \frac{9}{x} & \text { if } x

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

Explanation:

Step1: Recall the definition of the derivative

The derivative of a function \(f(x)\) at \(x = 0\) is given by \(f^{\prime}(0)=\lim_{x
ightarrow0}\frac{f(x)-f(0)}{x - 0}\).
Since \(f(0) = 0\) and \(f(x)=x\sin\frac{9}{x}\) for \(x
eq0\), then \(\frac{f(x)-f(0)}{x - 0}=\frac{x\sin\frac{9}{x}-0}{x}=\sin\frac{9}{x}\) for \(x
eq0\).

Step2: Evaluate the limit

We need to find \(\lim_{x
ightarrow0}\sin\frac{9}{x}\).
Let \(t=\frac{9}{x}\). As \(x
ightarrow0\), \(t
ightarrow\pm\infty\).
We know that the function \(y = \sin t\) oscillates between \(- 1\) and \(1\) as \(t
ightarrow\pm\infty\).
\(\lim_{x
ightarrow0}\sin\frac{9}{x}\) does not exist because \(\sin\frac{9}{x}\) oscillates infinitely as \(x
ightarrow0\).

Answer:

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