Let `f(x) = {{:(a sin^(2n)x,"for",x ge 0 and n rarr oo),(b cos^(2m)x - 1,"for",x lt 0 and m rarr oo):}` then

A. `f(0^(-)) ne f(0^(+))`
B. `f(0^(+)) ne f(0)`
C. `f(0^(-)) = f(0)`
D. `f(0^(-)) = f(0)`

Options

A. `f(0^(-)) ne f(0^(+))`
B. `f(0^(+)) ne f(0)`
C. `f(0^(-)) = f(0)`
D. f is continuous at x = 0

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined piecewise as follows:

\[ f(x) = \begin{cases} a \sin^{2n}(x) & \text{for } x \geq 0 \text{ and } n \to \infty \\ b \cos^{2m}(x) - 1 & \text{for } x < 0 \text{ and } m \to \infty \end{cases} \]

We need to evaluate the limits as \( x \) approaches 0 from the left (\( 0^- \)) and from the right (\( 0^+ \)), and also find the value of \( f(0) \).

### Step 1: Calculate \( f(0^-) \)

For \( x < 0 \), we have:

\[ f(x) = b \cos^{2m}(x) - 1 \]

As \( x \) approaches 0 from the left, we can substitute \( x = 0 \):

\[ f(0^-) = b \cos^{2m}(0) - 1 \]

Since \( \cos(0) = 1 \):

\[ f(0^-) = b \cdot 1^{2m} - 1 = b - 1 \]

### Step 2: Calculate \( f(0^+) \)

For \( x \geq 0 \), we have:

\[ f(x) = a \sin^{2n}(x) \]

As \( x \) approaches 0 from the right, we substitute \( x = 0 \):

\[ f(0^+) = a \sin^{2n}(0) \]

Since \( \sin(0) = 0 \):

\[ f(0^+) = a \cdot 0^{2n} = 0 \]

### Step 3: Evaluate \( f(0) \)

Since \( f(0) \) falls under the condition for \( x \geq 0 \):

\[ f(0) = a \sin^{2n}(0) = 0 \]

### Step 4: Compare the values

Now we compare the values we calculated:

- \( f(0^-) = b - 1 \)
- \( f(0^+) = 0 \)
- \( f(0) = 0 \)

### Conclusion

We can now analyze the options:

- \( f(0^-) \neq f(0^+) \) if \( b - 1 \neq 0 \) (which means \( b \neq 1 \)).
- \( f(0^+) = f(0) = 0 \).

Thus, the correct answer is:

**A. \( f(0^-) \neq f(0^+) \)**