Applying the L.Hospital rule , find the limits of the following functions :
`lim_(xto0) (ln(1+x^(2)))/(cos3x-e^(-x))`

Answer

To find the limit

\[ \lim_{x \to 0} \frac{\ln(1+x^2)}{\cos(3x) - e^{-x}}, \]

we will apply L'Hospital's Rule since the limit results in the indeterminate form \( \frac{0}{0} \).

### Step 1: Check the form of the limit

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

- The numerator:
\[ \ln(1 + 0^2) = \ln(1) = 0. \]

- The denominator:
\[ \cos(3 \cdot 0) - e^{-0} = \cos(0) - 1 = 1 - 1 = 0. \]

Since both the numerator and denominator approach 0, we have the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hospital's Rule

According to L'Hospital's Rule, we differentiate the numerator and the denominator separately.

- Differentiate the numerator:
\[ \frac{d}{dx} \ln(1+x^2) = \frac{1}{1+x^2} \cdot \frac{d}{dx}(1+x^2) = \frac{1}{1+x^2} \cdot 2x = \frac{2x}{1+x^2}. \]

- Differentiate the denominator:
\[ \frac{d}{dx}(\cos(3x) - e^{-x}) = -3\sin(3x) - (-e^{-x}) = -3\sin(3x) + e^{-x}. \]

### Step 3: Rewrite the limit

Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to 0} \frac{\frac{2x}{1+x^2}}{-3\sin(3x) + e^{-x}}. \]

### Step 4: Substitute \( x = 0 \) again

Now we substitute \( x = 0 \) into the new limit:

- The new numerator:
\[ \frac{2 \cdot 0}{1 + 0^2} = 0. \]

- The new denominator:
\[ -3\sin(3 \cdot 0) + e^{-0} = -3 \cdot 0 + 1 = 1. \]

Thus, we have:

\[ \frac{0}{1} = 0. \]

### Final Answer

Therefore, the limit is:

\[ \lim_{x \to 0} \frac{\ln(1+x^2)}{\cos(3x) - e^{-x}} = 0. \]