What is wrong with the following application of L'Hopital's rule ?
`lim_( xto 0) (1-3x)^(4//x)`

Answer

To analyze the limit \( \lim_{x \to 0} (1 - 3x)^{\frac{4}{x}} \) and determine what is wrong with the application of L'Hôpital's rule, we will follow these steps:

### Step 1: Identify the form of the limit
First, we need to evaluate the limit as \( x \) approaches 0.

\[ \lim_{x \to 0} (1 - 3x)^{\frac{4}{x}} \]

Substituting \( x = 0 \):

\[ (1 - 3 \cdot 0)^{\frac{4}{0}} = 1^{\infty} \]

This is an indeterminate form of type \( 1^{\infty} \).

### Step 2: Recognize that L'Hôpital's rule is not applicable
L'Hôpital's rule is only applicable for forms \( 0/0 \) or \( \infty/\infty \). Since we have \( 1^{\infty} \), we cannot apply L'Hôpital's rule directly.

### Step 3: Rewrite the limit using the exponential function
To resolve the \( 1^{\infty} \) form, we can rewrite the expression using the exponential function:

\[ (1 - 3x)^{\frac{4}{x}} = e^{\frac{4}{x} \ln(1 - 3x)} \]

### Step 4: Evaluate the limit of the exponent
Now we need to evaluate the limit of the exponent:

\[ \lim_{x \to 0} \frac{4}{x} \ln(1 - 3x) \]

### Step 5: Use the Taylor series expansion for \( \ln(1 - u) \)
Using the Taylor series expansion for \( \ln(1 - u) \) around \( u = 0 \):

\[ \ln(1 - 3x) \approx -3x \quad \text{as } x \to 0 \]

Thus,

\[ \lim_{x \to 0} \frac{4}{x} \ln(1 - 3x) \approx \lim_{x \to 0} \frac{4}{x} (-3x) = -12 \]

### Step 6: Substitute back into the exponential function
Now substituting back into the exponential function:

\[ \lim_{x \to 0} (1 - 3x)^{\frac{4}{x}} = e^{-12} \]

### Final Answer
Thus, the limit evaluates to:

\[ \lim_{x \to 0} (1 - 3x)^{\frac{4}{x}} = e^{-12} \]

### Summary of What Went Wrong
The application of L'Hôpital's rule was incorrect because the limit was not in the form \( 0/0 \) or \( \infty/\infty \), but rather \( 1^{\infty} \). This requires a different approach, such as rewriting the expression using the exponential function.