`lim_(x rarr 0)((5x^2+1)/(3x^2+1))^(1//x^2)`

Options

A. `e^2`
B. e
C. `e^-1`
D. none of these

Answer

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

### Step 1: Identify the form of the limit
As \( x \to 0 \), both the numerator and denominator approach 1:
\[ \frac{5x^2 + 1}{3x^2 + 1} \to \frac{1}{1} = 1 \]
Thus, we have the indeterminate form \( 1^{\infty} \).

**Hint:** Recognize indeterminate forms like \( 1^{\infty} \) can be resolved using logarithmic properties.

### Step 2: Rewrite the limit using the exponential function
We can rewrite the limit in a form that allows us to use the exponential function:
\[ \lim_{x \to 0} \left( \frac{5x^2 + 1}{3x^2 + 1} \right)^{\frac{1}{x^2}} = e^{\lim_{x \to 0} \frac{1}{x^2} \left( \frac{5x^2 + 1}{3x^2 + 1} - 1 \right)} \]

**Hint:** Use the property \( a^b = e^{b \ln(a)} \) to transform the limit.

### Step 3: Simplify the expression inside the limit
Now we need to simplify \( \frac{5x^2 + 1}{3x^2 + 1} - 1 \):
\[ \frac{5x^2 + 1}{3x^2 + 1} - 1 = \frac{(5x^2 + 1) - (3x^2 + 1)}{3x^2 + 1} = \frac{5x^2 - 3x^2}{3x^2 + 1} = \frac{2x^2}{3x^2 + 1} \]

**Hint:** Combine fractions by finding a common denominator.

### Step 4: Substitute back into the limit
Now substitute this back into the limit:
\[ \lim_{x \to 0} \frac{1}{x^2} \left( \frac{2x^2}{3x^2 + 1} \right) = \lim_{x \to 0} \frac{2}{3x^2 + 1} \]

**Hint:** Simplify the limit by substituting \( x = 0 \).

### Step 5: Evaluate the limit
Now evaluate the limit:
\[ \lim_{x \to 0} \frac{2}{3x^2 + 1} = \frac{2}{3(0) + 1} = 2 \]

**Hint:** Direct substitution works well when the limit is not indeterminate.

### Step 6: Final result
Now substitute back into the exponential function:
\[ e^{\lim_{x \to 0} \frac{1}{x^2} \left( \frac{5x^2 + 1}{3x^2 + 1} - 1 \right)} = e^2 \]

Thus, the final answer is:
\[ \lim_{x \to 0} \left( \frac{5x^2 + 1}{3x^2 + 1} \right)^{\frac{1}{x^2}} = e^2 \]

### Final Answer
\[ \boxed{e^2} \]