If `f(x)={((a^(2[x]+{x})-1)"/ "(2[x]+{x}),x!=0),(log_(e)a,x=0):}`
Then at x=0, (where [x] and {x} are the greatest integer function and fraction part of x respectively

Options

A. `f(x)` is continuous
B. `f(x)` is discontinuous
C. `lim_(xto0^(+))f(x)=a`
D. `lim_(xto0)f(x)=1/(log_(e)a)`

Answer

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

\[ f(x) = \begin{cases} \frac{a^{2[x] + \{x\}} - 1}{2[x] + \{x\}}, & x \neq 0 \\ \log_e a, & x = 0 \end{cases} \]

We need to determine the behavior of \( f(x) \) as \( x \) approaches 0 from both sides (left-hand limit and right-hand limit).

### Step 1: Calculate the Left-Hand Limit as \( x \to 0^- \)

For \( x \to 0^- \), we have:

- \( [x] = -1 \) (the greatest integer less than or equal to \( x \))
- \( \{x\} = x - [x] = x + 1 \)

Thus, we can rewrite \( f(x) \):

\[ f(x) = \frac{a^{2(-1) + (x + 1)} - 1}{2(-1) + (x + 1)} = \frac{a^{-2 + x + 1} - 1}{-2 + x + 1} = \frac{a^{x - 1} - 1}{x - 1} \]

Now we need to compute the limit:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{a^{x - 1} - 1}{x - 1} \]

### Step 2: Apply L'Hôpital's Rule

As \( x \to 0^- \), both the numerator and denominator approach 0, so we can apply L'Hôpital's Rule:

\[ \text{Differentiate the numerator: } \frac{d}{dx}(a^{x - 1}) = a^{x - 1} \ln a \]
\[ \text{Differentiate the denominator: } \frac{d}{dx}(x - 1) = 1 \]

Thus, we have:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{a^{x - 1} \ln a}{1} = a^{-1} \ln a \]

### Step 3: Calculate the Right-Hand Limit as \( x \to 0^+ \)

For \( x \to 0^+ \):

- \( [x] = 0 \)
- \( \{x\} = x \)

Thus, we can rewrite \( f(x) \):

\[ f(x) = \frac{a^{2(0) + x} - 1}{2(0) + x} = \frac{a^{x} - 1}{x} \]

Now we compute the limit:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{a^{x} - 1}{x} \]

### Step 4: Apply L'Hôpital's Rule Again

As \( x \to 0^+ \), both the numerator and denominator approach 0, so we apply L'Hôpital's Rule:

\[ \text{Differentiate the numerator: } \frac{d}{dx}(a^{x}) = a^{x} \ln a \]
\[ \text{Differentiate the denominator: } \frac{d}{dx}(x) = 1 \]

Thus, we have:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{a^{x} \ln a}{1} = \ln a \]

### Step 5: Compare the Limits

Now we compare the left-hand limit and the right-hand limit:

- Left-hand limit: \( \frac{\ln a}{a} \)
- Right-hand limit: \( \ln a \)

Since \( \frac{\ln a}{a} \neq \ln a \) for \( a \neq 1 \), we conclude that the limits are not equal.

### Conclusion

Since the left-hand limit does not equal the right-hand limit, the function \( f(x) \) is discontinuous at \( x = 0 \).

### Final Answer

The function \( f(x) \) is discontinuous at \( x = 0 \).

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