The function f defined as `f(x) = (sin x^(2))//x` for `x ne 0` and `f(0)=0` is:

Options

A. continuous and derivable at x = 0,
B. neither continuous nor derivable at x = 0,
C. continuous but not derivable at x = 0
D. none of these.

Answer

To determine the continuity and differentiability of the function \( f(x) = \frac{\sin(x^2)}{x} \) for \( x \neq 0 \) and \( f(0) = 0 \), we will follow these steps:

### Step 1: Check Continuity at \( x = 0 \)

To check the continuity of \( f \) at \( x = 0 \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0 and compare it with \( f(0) \).

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin(x^2)}{x} \]

Using the substitution \( h = x^2 \), as \( x \to 0 \), \( h \to 0 \) as well. Thus, we can rewrite the limit:

\[ \lim_{x \to 0} \frac{\sin(x^2)}{x} = \lim_{h \to 0} \frac{\sin(h)}{\sqrt{h}} = \lim_{h \to 0} \frac{\sin(h)}{h} \cdot \frac{h}{\sqrt{h}} = \lim_{h \to 0} \frac{\sin(h)}{h} \cdot \sqrt{h} \]

We know that \( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \). Therefore, we have:

\[ \lim_{h \to 0} \sqrt{h} = 0 \]

Thus,

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

Since \( f(0) = 0 \), we find that:

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

This shows that \( f \) is continuous at \( x = 0 \).

### Step 2: Check Differentiability at \( x = 0 \)

To check differentiability at \( x = 0 \), we need to compute the left-hand and right-hand derivatives.

**Left-hand derivative:**

\[ f'(0) = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{-h} = \lim_{h \to 0} \frac{f(-h)}{-h} = \lim_{h \to 0} \frac{\sin((-h)^2)}{-h} = \lim_{h \to 0} \frac{\sin(h^2)}{-h} \]

Using the same substitution as before:

\[ = \lim_{h \to 0} \frac{\sin(h^2)}{h} = \lim_{h \to 0} \frac{\sin(h^2)}{h^2} \cdot h = \lim_{h \to 0} \frac{\sin(h^2)}{h^2} \cdot h \]

Since \( \lim_{h \to 0} \frac{\sin(h^2)}{h^2} = 1 \):

\[ \lim_{h \to 0} h = 0 \]

Thus, the left-hand derivative at \( x = 0 \) is 0.

**Right-hand derivative:**

\[ f'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h)}{h} = \lim_{h \to 0} \frac{\sin(h^2)}{h^2} \]

Again, we have:

\[ \lim_{h \to 0} \frac{\sin(h^2)}{h^2} = 1 \]

Thus, the right-hand derivative at \( x = 0 \) is also 0.

Since both the left-hand and right-hand derivatives at \( x = 0 \) are equal, we conclude that \( f \) is differentiable at \( x = 0 \).

### Conclusion

The function \( f(x) \) is both continuous and differentiable at \( x = 0 \).

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