Evaluate the following limits :
`lim_(x to 0)(x(e^(2+x)-e^(2)))/(1-cosx)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{x(e^{2+x} - e^2)}{1 - \cos x}, \]

we first check the form of the limit by substituting \(x = 0\):

- The numerator becomes \(0(e^{2+0} - e^2) = 0(1 - 1) = 0\).
- The denominator becomes \(1 - \cos(0) = 1 - 1 = 0\).

Since we have a \( \frac{0}{0} \) indeterminate form, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \), we can differentiate the numerator and the denominator separately.

### Step 1: Differentiate the numerator and denominator

The numerator is \(x(e^{2+x} - e^2)\). We will use the product rule to differentiate it:

Let \(u = x\) and \(v = e^{2+x} - e^2\).

Then,

\[ \frac{du}{dx} = 1, \]
\[ \frac{dv}{dx} = e^{2+x} \cdot \frac{d}{dx}(2+x) = e^{2+x} \cdot 1 = e^{2+x}. \]

Using the product rule:

\[ \frac{d}{dx}[u \cdot v] = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} = x \cdot e^{2+x} + (e^{2+x} - e^2) \cdot 1. \]

Thus, the derivative of the numerator is:

\[ x e^{2+x} + (e^{2+x} - e^2). \]

Now, simplifying this gives:

\[ x e^{2+x} + e^{2+x} - e^2 = (x + 1)e^{2+x} - e^2. \]

The derivative of the denominator \(1 - \cos x\) is:

\[ \frac{d}{dx}[1 - \cos x] = \sin x. \]

### Step 2: Rewrite the limit using L'Hôpital's Rule

Now we can rewrite the limit:

\[ \lim_{x \to 0} \frac{(x + 1)e^{2+x} - e^2}{\sin x}. \]

### Step 3: Substitute \(x = 0\) again

Substituting \(x = 0\):

- The numerator becomes \((0 + 1)e^{2+0} - e^2 = e^2 - e^2 = 0\).
- The denominator becomes \(\sin(0) = 0\).

We again have a \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again.

### Step 4: Differentiate again

Now we differentiate the numerator and denominator again.

For the numerator \((x + 1)e^{2+x} - e^2\):

Using the product rule again:

Let \(u = x + 1\) and \(v = e^{2+x}\).

Then,

\[ \frac{du}{dx} = 1, \]
\[ \frac{dv}{dx} = e^{2+x}. \]

Using the product rule:

\[ \frac{d}{dx}[u \cdot v] = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} = (x + 1)e^{2+x} + e^{2+x} \cdot 1 = (x + 2)e^{2+x}. \]

The derivative of the constant \(e^2\) is \(0\).

Thus, the derivative of the numerator is:

\[ (x + 2)e^{2+x}. \]

The derivative of the denominator \(\sin x\) is \(\cos x\).

### Step 5: Rewrite the limit again

Now we have:

\[ \lim_{x \to 0} \frac{(x + 2)e^{2+x}}{\cos x}. \]

### Step 6: Substitute \(x = 0\) again

Substituting \(x = 0\):

- The numerator becomes \((0 + 2)e^{2+0} = 2e^2\).
- The denominator becomes \(\cos(0) = 1\).

Thus, the limit evaluates to:

\[ \frac{2e^2}{1} = 2e^2. \]

### Final Answer

The limit is

\[ \boxed{2e^2}. \]