Evaluate `lim_(xto0) (e^(x)-e^(-x)-2x)/(x-sinx).`

Explanation

To evaluate the limit

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

we will follow these steps:

### Step 1: Substitute \( x = 0 \)

First, we substitute \( x = 0 \) into the expression to check if it results in an indeterminate form.

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

This gives us the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply Taylor Series Expansion

Since we have an indeterminate form, we can use the Taylor series expansions for \( e^x \) and \( \sin x \) around \( x = 0 \).

The Taylor series expansion for \( e^x \) is:

\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]

The Taylor series expansion for \( e^{-x} \) is:

\[ e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots \]

Subtracting these two expansions:

\[ e^x - e^{-x} = \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\right) - \left(1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots\right) \]

This simplifies to:

\[ e^x - e^{-x} = 2x + \frac{2x^3}{3!} + \cdots \]

### Step 3: Substitute in the Numerator

Now we substitute this result into the numerator of our limit:

\[ e^x - e^{-x} - 2x = \left(2x + \frac{2x^3}{3!} + \cdots\right) - 2x = \frac{2x^3}{3!} + \cdots \]

### Step 4: Expand \( \sin x \)

Next, we expand \( \sin x \):

\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \]

Now substituting this into the denominator:

\[ x - \sin x = x - \left(x - \frac{x^3}{3!} + \cdots\right) = \frac{x^3}{3!} - \frac{x^5}{5!} + \cdots \]

### Step 5: Substitute into the Limit

Now we substitute our results back into the limit:

\[ \lim_{x \to 0} \frac{\frac{2x^3}{3!} + \cdots}{\frac{x^3}{3!} - \frac{x^5}{5!} + \cdots} \]

### Step 6: Factor out \( x^3 \)

Factoring out \( x^3 \) from both the numerator and denominator:

\[ = \lim_{x \to 0} \frac{2 + \cdots}{1 - \frac{x^2}{5!} + \cdots} \]

### Step 7: Evaluate the Limit

As \( x \to 0 \), the higher-order terms vanish:

\[ = \frac{2}{1} = 2 \]

### Final Answer

Thus, the limit is:

\[ \boxed{2} \]