Applying the L.Hospital rule , find the limits of the following functions :
`lim_(xto0) (e^(x)-e^(-x)-2x)/(x-sin x)`

Answer

To find the limit of the function

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

we will apply L'Hospital's rule since the limit results in the indeterminate form \( \frac{0}{0} \).

### Step 1: Check the form of the limit
First, we evaluate the numerator and denominator at \( x = 0 \):

- Numerator:
\[ e^0 - e^{-0} - 2(0) = 1 - 1 - 0 = 0 \]

- Denominator:
\[ 0 - \sin(0) = 0 - 0 = 0 \]

Since both the numerator and denominator approach 0, we have the indeterminate form \( \frac{0}{0} \).

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

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

### Step 3: Evaluate the new limit
Now we check the new limit at \( x = 0 \):

- Numerator:
\[ e^0 + e^{-0} - 2 = 1 + 1 - 2 = 0 \]

- Denominator:
\[ 1 - \cos(0) = 1 - 1 = 0 \]

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

### Step 4: Apply L'Hospital's Rule again
Differentiate the numerator and 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: Evaluate the new limit
Now we check the limit at \( x = 0 \):

- Numerator:
\[ e^0 - e^{-0} = 1 - 1 = 0 \]

- Denominator:
\[ \sin(0) = 0 \]

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

### Step 6: Apply L'Hospital's Rule again
Differentiate the numerator and 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: Evaluate the new limit
Now we check the limit at \( x = 0 \):

- Numerator:
\[ e^0 + e^{-0} = 1 + 1 = 2 \]

- Denominator:
\[ \cos(0) = 1 \]

Thus, the limit is:
\[ \frac{2}{1} = 2 \]

### Final Answer
The limit is:
\[ \lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x} = 2 \]