Prove that `f(x) = {sinx/x ; x != 0 and 1 ; x=0`. is continuous at `x=0`.

Answer

To prove that the function

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

is continuous at \(x = 0\), we need to show that

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

### Step 1: Identify \(f(0)\)

From the definition of the function, we have:

\[ f(0) = 1. \]

### Step 2: Calculate \(\lim_{x \to 0} f(x)\)

For \(x \neq 0\), the function is given by \(f(x) = \frac{\sin x}{x}\). We need to find the limit as \(x\) approaches 0:

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

### Step 3: Use the standard limit result

It is a well-known limit that:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1. \]

### Step 4: Conclude the limit

Thus, we have:

\[ \lim_{x \to 0} f(x) = 1. \]

### Step 5: Compare the limit with \(f(0)\)

Now we compare the limit with the value of the function at 0:

\[ \lim_{x \to 0} f(x) = 1 = f(0). \]

### Conclusion

Since

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

we conclude that \(f(x)\) is continuous at \(x = 0\).

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