Using L'Hospital's rule, evaluate : `lim_(xrarr0)((e^(x)-e^(-x)-2x))/(x-"sin"x)`

Answer

To evaluate the limit using L'Hospital's rule, we start with the expression:

\[ \lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x} \]

### Step 1: Check the form of the limit
First, we substitute \(x = 0\) into the expression:

- The numerator becomes:
\[ e^0 - e^0 - 2 \cdot 0 = 1 - 1 - 0 = 0 \]

- The denominator becomes:
\[ 0 - \sin(0) = 0 - 0 = 0 \]

Since both the numerator and denominator approach 0, we have the indeterminate form \( \frac{0}{0} \). Thus, we can apply L'Hospital's rule.

### Step 2: Apply L'Hospital's Rule
According to L'Hospital's rule, we differentiate the numerator and the denominator separately:

- Differentiate the numerator:
\[ \frac{d}{dx}(e^x - e^{-x} - 2x) = e^x + e^{-x} - 2 \]

- Differentiate the denominator:
\[ \frac{d}{dx}(x - \sin x) = 1 - \cos x \]

Now, we rewrite the limit:
\[ \lim_{x \to 0} \frac{e^x + e^{-x} - 2}{1 - \cos x} \]

### Step 3: Substitute \(x = 0\) again
We substitute \(x = 0\) into the new expression:

- The numerator becomes:
\[ e^0 + e^0 - 2 = 1 + 1 - 2 = 0 \]

- The denominator becomes:
\[ 1 - \cos(0) = 1 - 1 = 0 \]

Again, we have the indeterminate form \( \frac{0}{0} \). We apply L'Hospital's rule again.

### Step 4: Differentiate again
Differentiate the numerator and the denominator again:

- Differentiate the numerator:
\[ \frac{d}{dx}(e^x + e^{-x} - 2) = e^x - e^{-x} \]

- Differentiate the denominator:
\[ \frac{d}{dx}(1 - \cos x) = \sin x \]

Now, we rewrite the limit:
\[ \lim_{x \to 0} \frac{e^x - e^{-x}}{\sin x} \]

### Step 5: Substitute \(x = 0\) again
Substituting \(x = 0\):

- The numerator becomes:
\[ e^0 - e^0 = 1 - 1 = 0 \]

- The denominator becomes:
\[ \sin(0) = 0 \]

We still have the form \( \frac{0}{0} \), so we apply L'Hospital's rule one more time.

### Step 6: Differentiate once more
Differentiate the numerator and the denominator again:

- Differentiate the numerator:
\[ \frac{d}{dx}(e^x - e^{-x}) = e^x + e^{-x} \]

- Differentiate the denominator:
\[ \frac{d}{dx}(\sin x) = \cos x \]

Now, we rewrite the limit:
\[ \lim_{x \to 0} \frac{e^x + e^{-x}}{\cos x} \]

### Step 7: Substitute \(x = 0\) one last time
Substituting \(x = 0\):

- The numerator becomes:
\[ e^0 + e^0 = 1 + 1 = 2 \]

- The denominator becomes:
\[ \cos(0) = 1 \]

Thus, we have:
\[ \lim_{x \to 0} \frac{2}{1} = 2 \]

### Final Answer
The limit is:
\[ \boxed{2} \]