Evaluate the following limits :
`lim_(x to 0)(e^(x)-e^(sinx))/(x-sinx)`

Answer

To evaluate the limit

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

we will follow these steps:

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

First, we substitute \(x = 0\) into the limit:

\[ \frac{e^0 - e^{\sin 0}}{0 - \sin 0} = \frac{1 - 1}{0 - 0} = \frac{0}{0}. \]

This is an indeterminate form, so we need to manipulate the expression further.

**Hint:** When you encounter the form \(0/0\), consider using algebraic manipulation or L'Hôpital's rule.

### Step 2: Rewrite the expression

We can rewrite the expression by factoring out \(e^{\sin x}\) from the numerator:

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

Thus, we can rewrite the limit as:

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

**Hint:** Factoring out common terms can simplify the limit evaluation.

### Step 3: Analyze \(x - \sin x\)

Next, we need to analyze \(x - \sin x\). We know that as \(x \to 0\), \(x - \sin x\) can be approximated using the Taylor series expansion of \(\sin x\):

\[ \sin x \approx x - \frac{x^3}{6} + O(x^5). \]

Thus,

\[ x - \sin x \approx \frac{x^3}{6}. \]

**Hint:** Using Taylor series can help in simplifying expressions near a point.

### Step 4: Substitute back into the limit

Now, substituting back, we have:

\[ \lim_{x \to 0} \frac{e^{\sin x}(e^{x - \sin x} - 1)}{\frac{x^3}{6}} = \lim_{x \to 0} \frac{6e^{\sin x}(e^{x - \sin x} - 1)}{x^3}. \]

**Hint:** When rewriting the limit, ensure that you maintain the correct form of the expression.

### Step 5: Evaluate \(e^{x - \sin x}\)

Next, we need to evaluate \(e^{x - \sin x}\). As \(x \to 0\), \(x - \sin x \to 0\), and we can use the fact that:

\[ e^u - 1 \approx u \text{ for small } u. \]

Thus,

\[ e^{x - \sin x} - 1 \approx (x - \sin x). \]

**Hint:** Recognizing limits of exponential functions can simplify calculations.

### Step 6: Substitute and simplify

Substituting this back into our limit gives:

\[ \lim_{x \to 0} \frac{6e^{\sin x}(x - \sin x)}{x^3}. \]

Now, we can replace \(x - \sin x\) with \(\frac{x^3}{6}\):

\[ \lim_{x \to 0} \frac{6e^{\sin x} \cdot \frac{x^3}{6}}{x^3} = \lim_{x \to 0} e^{\sin x}. \]

**Hint:** Simplifying the limit to a single term can often lead to an easier evaluation.

### Step 7: Evaluate the final limit

Finally, as \(x \to 0\), \(\sin x \to 0\), so:

\[ e^{\sin x} \to e^0 = 1. \]

Thus, the limit evaluates to:

\[ \lim_{x \to 0} \frac{e^x - e^{\sin x}}{x - \sin x} = 1. \]

### Final Answer

The final result is:

\[ \boxed{1}. \]