The value of `lim_(x rarr 0) ((1+5x^(2))/(1+3x^(2)))^(1/x^2)` is

Options

A. `e^(2)`
B. e
C. 1/e
D. `1//e^(2)`

Answer

To solve the limit \( \lim_{x \to 0} \left( \frac{1 + 5x^2}{1 + 3x^2} \right)^{\frac{1}{x^2}} \), we can follow these steps:

### Step 1: Rewrite the limit
We start by rewriting the limit in a more manageable form:
\[ L = \lim_{x \to 0} \left( \frac{1 + 5x^2}{1 + 3x^2} \right)^{\frac{1}{x^2}} \]

### Step 2: Take the natural logarithm
To simplify the expression, we take the natural logarithm:
\[ \ln L = \lim_{x \to 0} \frac{1}{x^2} \ln \left( \frac{1 + 5x^2}{1 + 3x^2} \right) \]

### Step 3: Simplify the logarithm
We can simplify the logarithm:
\[ \ln \left( \frac{1 + 5x^2}{1 + 3x^2} \right) = \ln(1 + 5x^2) - \ln(1 + 3x^2) \]
Using the Taylor expansion \( \ln(1 + u) \approx u \) when \( u \) is small, we have:
\[ \ln(1 + 5x^2) \approx 5x^2 \quad \text{and} \quad \ln(1 + 3x^2) \approx 3x^2 \]
Thus,
\[ \ln \left( \frac{1 + 5x^2}{1 + 3x^2} \right) \approx 5x^2 - 3x^2 = 2x^2 \]

### Step 4: Substitute back into the limit
Now substituting back into the limit:
\[ \ln L = \lim_{x \to 0} \frac{1}{x^2} (2x^2) = \lim_{x \to 0} 2 = 2 \]

### Step 5: Exponentiate to find L
Now, exponentiate both sides to solve for \( L \):
\[ L = e^{\ln L} = e^2 \]

### Final Answer
Thus, the value of the limit is:
\[ \lim_{x \to 0} \left( \frac{1 + 5x^2}{1 + 3x^2} \right)^{\frac{1}{x^2}} = e^2 \]