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

Answer

To evaluate the limit

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

we first observe that substituting \( x = 0 \) directly into the expression gives us:

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

which is an indeterminate form. Therefore, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form of type \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can take the derivative of the numerator and the denominator.

### Step 1: Differentiate the numerator and denominator

1. **Numerator:** The derivative of \( e^x - e^{-x} \) is:
\[ \frac{d}{dx}(e^x - e^{-x}) = e^x + e^{-x}. \]

2. **Denominator:** The derivative of \( \sin x \) is:
\[ \frac{d}{dx}(\sin x) = \cos x. \]

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

Now we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{e^x + e^{-x}}{\cos x}. \]

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

Substituting \( x = 0 \) into the new expression gives us:

\[ \frac{e^0 + e^{-0}}{\cos 0} = \frac{1 + 1}{1} = \frac{2}{1} = 2. \]

### Conclusion

Thus, the limit evaluates to:

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