Evaluate the limits, if exist `lim_(x rarr 0)(e^(4x)-1)/x`.

Explanation

To evaluate the limit \( \lim_{x \to 0} \frac{e^{4x} - 1}{x} \), we can follow these steps:

### Step 1: Identify the form of the limit
As \( x \to 0 \), both the numerator \( e^{4x} - 1 \) and the denominator \( x \) approach 0. This gives us the indeterminate form \( \frac{0}{0} \).

### Step 2: Use L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

In this case, let \( f(x) = e^{4x} - 1 \) and \( g(x) = x \).

### Step 3: Differentiate the numerator and denominator
Now we differentiate \( f(x) \) and \( g(x) \):

- The derivative of \( f(x) = e^{4x} - 1 \) is \( f'(x) = 4e^{4x} \).
- The derivative of \( g(x) = x \) is \( g'(x) = 1 \).

### Step 4: Apply L'Hôpital's Rule
Now we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{e^{4x} - 1}{x} = \lim_{x \to 0} \frac{4e^{4x}}{1} \]

### Step 5: Evaluate the limit
Now we evaluate the limit as \( x \to 0 \):

\[ \lim_{x \to 0} 4e^{4x} = 4e^{0} = 4 \cdot 1 = 4 \]

### Conclusion
Thus, the limit exists and is equal to:

\[ \lim_{x \to 0} \frac{e^{4x} - 1}{x} = 4 \]