Discuss the continuity of `f(x) ={:{(sin""(1)/(x)", "x ne 0),(" "1", " x= 0):},`

Answer

To discuss the continuity of the function

\[ f(x) = \begin{cases} \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \]

we need to check if the function is continuous at \(x = 0\) and for \(x \neq 0\).

### Step 1: Check the value of the function at \(x = 0\)

We have:

\[ f(0) = 1 \]

### Step 2: Find the limit of \(f(x)\) as \(x\) approaches 0

We need to evaluate:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \sin\left(\frac{1}{x}\right) \]

### Step 3: Analyze the limit

As \(x\) approaches 0, \(\frac{1}{x}\) approaches infinity. The sine function oscillates between -1 and 1 for all real numbers. Therefore,

\[ \sin\left(\frac{1}{x}\right) \]

does not approach a single value as \(x\) approaches 0. Instead, it oscillates indefinitely between -1 and 1.

### Step 4: Conclusion about the limit

Since the limit does not exist, we can conclude that:

\[ \lim_{x \to 0} f(x) \text{ does not exist.} \]

### Step 5: Check continuity at \(x = 0\)

For \(f(x)\) to be continuous at \(x = 0\), the following must hold:

\[ \lim_{x \to 0} f(x) = f(0) \]

Since \(\lim_{x \to 0} f(x)\) does not exist, it cannot equal \(f(0) = 1\). Thus, \(f(x)\) is not continuous at \(x = 0\).

### Final Conclusion

The function \(f(x)\) is continuous for all \(x \neq 0\) but is not continuous at \(x = 0\).

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