Let `f(x) = (e^x x cosx-x log_e(1+x)-x)/x^2, x!=0.` If `f(x)` is continuous at `x = 0,` then `f(0)` is equal to

Options

A. 0
B. 1
C. `-1`
D. 2

Answer

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

\[ f(x) = \frac{e^x x \cos x - x \log(1+x) - x}{x^2}, \quad x \neq 0, \]

we need to ensure that \( f(x) \) is continuous at \( x = 0 \). This requires us to find the limit of \( f(x) \) as \( x \) approaches 0.

### Step 1: Expand the functions using Taylor series

We will use the Taylor series expansions for \( e^x \), \( \cos x \), and \( \log(1+x) \):

1. **For \( e^x \)**:
\[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4) \]

2. **For \( \cos x \)**:
\[ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6) \]

3. **For \( \log(1+x) \)**:
\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + O(x^5) \]

### Step 2: Substitute the expansions into \( f(x) \)

Now, substituting these expansions into \( f(x) \):

\[ f(x) = \frac{\left(1 + x + \frac{x^2}{2} + O(x^3)\right) x \left(1 - \frac{x^2}{2} + O(x^4)\right) - x\left(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + O(x^5)\right) - x}{x^2} \]

### Step 3: Simplify the numerator

Calculating the first term:

\[ e^x x \cos x = \left(1 + x + \frac{x^2}{2} + O(x^3)\right) x \left(1 - \frac{x^2}{2} + O(x^4)\right) \]
\[ = x + x^2 + \frac{x^3}{2} - \frac{x^3}{2} + O(x^4) = x + x^2 + O(x^4) \]

Now for the second term:

\[ -x \log(1+x) = -x\left(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + O(x^5)\right) = -x^2 + \frac{x^3}{2} - \frac{x^4}{4} + O(x^5) \]

Putting these together:

\[ f(x) = \frac{x + x^2 + O(x^4) - (-x^2 + \frac{x^3}{2} - \frac{x^4}{4} + O(x^5)) - x}{x^2} \]
\[ = \frac{x + x^2 + x^2 - \frac{x^3}{2} + \frac{x^4}{4} + O(x^4) - x}{x^2} \]
\[ = \frac{2x^2 - \frac{x^3}{2} + O(x^4)}{x^2} \]
\[ = 2 - \frac{x}{2} + O(x^2) \]

### Step 4: Take the limit as \( x \to 0 \)

Now, we take the limit as \( x \) approaches 0:

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

### Step 5: Conclusion

Thus, for \( f(x) \) to be continuous at \( x = 0 \), we define:

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

### Final Answer

Therefore, \( f(0) = 2 \).

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