If `f(x)={{:((e^((1+(1)/(x)))-a)/(e^((1)/(x))+1),":",xne0),(e,":",x=0):}` (where a and b are arbitrary constants) is continuous at x = 0, then the value of `a^(2)` is equal to
(use e = 2.7)

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 0 \). The function is defined as follows:

\[ f(x) = \begin{cases} \frac{e^{(1 + \frac{1}{x})} - a}{e^{\frac{1}{x}} + 1} & \text{if } x \neq 0 \\ e & \text{if } x = 0 \end{cases} \]

### Step 1: Find the right-hand limit as \( x \) approaches 0.

We need to evaluate the limit:

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

### Step 2: Substitute \( x \) with \( 0 + h \) where \( h \to 0^+ \).

Let \( x = h \) where \( h \to 0^+ \):

\[ \lim_{h \to 0^+} \frac{e^{(1 + \frac{1}{h})} - a}{e^{\frac{1}{h}} + 1} \]

### Step 3: Analyze the exponential terms as \( h \to 0^+ \).

As \( h \to 0^+ \), \( \frac{1}{h} \to +\infty \). Therefore:

- \( e^{\frac{1}{h}} \to +\infty \)
- \( e^{(1 + \frac{1}{h})} = e \cdot e^{\frac{1}{h}} \to +\infty \)

Thus, we can rewrite the limit:

\[ \lim_{h \to 0^+} \frac{+\infty - a}{+\infty + 1} \]

### Step 4: Simplify the limit.

Since both the numerator and denominator approach infinity, we can apply L'Hôpital's Rule or analyze the dominant terms:

\[ \lim_{h \to 0^+} \frac{e^{(1 + \frac{1}{h})}}{e^{\frac{1}{h}}} = \lim_{h \to 0^+} e \cdot \frac{e^{\frac{1}{h}}}{e^{\frac{1}{h}}} = e \]

Thus, the limit simplifies to:

\[ \lim_{h \to 0^+} \frac{+\infty - a}{+\infty} = 1 \]

### Step 5: Set the limit equal to the function value at \( x = 0 \).

Since \( f(0) = e \), we have:

\[ 1 - \frac{a}{+\infty} = e \]

This means:

\[ -\frac{a}{+\infty} \to 0 \]

### Step 6: Solve for \( a \).

From the limit, we get:

\[ -a = e \implies a = -e \]

### Step 7: Find \( a^2 \).

Now, we need to find \( a^2 \):

\[ a^2 = (-e)^2 = e^2 \]

### Step 8: Substitute the value of \( e \).

Given \( e \approx 2.7 \):

\[ a^2 = (2.7)^2 = 7.29 \]

### Final Answer:

Thus, the value of \( a^2 \) is \( 7.29 \).

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