If `f(x)` is continuous at `x=0`, where `f(x)=(e^(x^(2))-cosx)/(x^(2))`, for `x!=0`, then `f(0)=`

Options

A. `3/2`
B. `1/2`
C. 1
D. `(-1)/(2)`

Answer

To find the value of \( f(0) \) for the function

\[ f(x) = \frac{e^{x^2} - \cos x}{x^2} \quad \text{for } x \neq 0, \]

we need to ensure that the function is continuous at \( x = 0 \). For continuity at \( x = 0 \), we need to find

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

### Step 1: Calculate the limit as \( x \) approaches 0

We need to evaluate

\[ \lim_{x \to 0} \frac{e^{x^2} - \cos x}{x^2}. \]

### Step 2: Substitute \( x = 0 \)

Substituting \( x = 0 \) directly into the function gives us an indeterminate form \( \frac{0}{0} \):

\[ e^{0^2} - \cos(0) = 1 - 1 = 0 \quad \text{and} \quad 0^2 = 0. \]

### Step 3: Apply L'Hôpital's Rule

Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

if the limit on the right exists. Here, we differentiate the numerator and the denominator.

- The derivative of the numerator \( e^{x^2} - \cos x \) is:

\[ \frac{d}{dx}(e^{x^2}) = 2x e^{x^2} \quad \text{and} \quad \frac{d}{dx}(-\cos x) = \sin x. \]

Thus, the derivative of the numerator is:

\[ 2x e^{x^2} + \sin x. \]

- The derivative of the denominator \( x^2 \) is:

\[ \frac{d}{dx}(x^2) = 2x. \]

### Step 4: Rewrite the limit using L'Hôpital's Rule

Now we can rewrite the limit:

\[ \lim_{x \to 0} \frac{2x e^{x^2} + \sin x}{2x}. \]

### Step 5: Simplify the expression

We can simplify this limit:

\[ \lim_{x \to 0} \left( \frac{2x e^{x^2}}{2x} + \frac{\sin x}{2x} \right) = \lim_{x \to 0} \left( e^{x^2} + \frac{\sin x}{2x} \right). \]

### Step 6: Evaluate the limit

Now we evaluate each part:

1. As \( x \to 0 \), \( e^{x^2} \to e^0 = 1 \).
2. The limit \( \frac{\sin x}{x} \to 1 \) as \( x \to 0 \), so \( \frac{\sin x}{2x} \to \frac{1}{2} \).

Putting it all together:

\[ \lim_{x \to 0} \left( e^{x^2} + \frac{\sin x}{2x} \right) = 1 + \frac{1}{2} = \frac{3}{2}. \]

### Conclusion

Thus, the value of \( f(0) \) is

\[ f(0) = \frac{3}{2}. \]

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