If `f(x)=sum_(n=0)^oo x^n/(n!)(loga)^n`, then at `x = 0, f(x)` (a) has no limit (b) is discontinuous (c) is continuous but not differentiable (d) is differentiable

Options

A. has no limit
B. is discontinuous
C. is continuous but not differentiable
D. is differentiable

Answer

To solve the problem, we need to analyze the function defined by the series:

\[ f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} (\log a)^n \]

This series resembles the Taylor series expansion for the exponential function.

### Step 1: Recognize the Series
The series can be rewritten using the exponential function:

\[ f(x) = \sum_{n=0}^{\infty} \frac{(x \log a)^n}{n!} = e^{x \log a} \]

### Step 2: Simplify the Function
Using the property of exponents, we can simplify \(e^{x \log a}\):

\[ f(x) = a^x \]

### Step 3: Evaluate at \(x = 0\)
Now, we need to evaluate \(f(x)\) at \(x = 0\):

\[ f(0) = a^0 = 1 \]

### Step 4: Analyze Continuity and Differentiability
The function \(f(x) = a^x\) is an exponential function, which is continuous and differentiable for all real numbers.

### Conclusion
Since \(f(x)\) is continuous and differentiable at \(x = 0\), we can conclude that the correct answer is:

**(d) is differentiable.**

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