Evaluate :
`lim_(x to 0 ) ((1+6x^(2))/(1+4x^(2)))^(1/(x^(2)))`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \left(\frac{1 + 6x^2}{1 + 4x^2}\right)^{\frac{1}{x^2}}, \]

we can follow these steps:

### Step 1: Identify the form of the limit
First, we substitute \(x = 0\) into the expression:

\[ \frac{1 + 6(0)^2}{1 + 4(0)^2} = \frac{1 + 0}{1 + 0} = \frac{1}{1} = 1. \]

Thus, we have the form \(1^{\infty}\), which is an indeterminate form.

### Step 2: Apply the logarithmic limit technique
We can rewrite the limit using the exponential function. We know that if \(L = \lim_{x \to a} f(x)^{g(x)}\) and \(f(x) \to 1\) and \(g(x) \to \infty\), then:

\[ L = e^{\lim_{x \to a} g(x) \cdot (f(x) - 1)}. \]

In our case, let \(f(x) = \frac{1 + 6x^2}{1 + 4x^2}\) and \(g(x) = \frac{1}{x^2}\).

### Step 3: Calculate \(f(x) - 1\)
Now we need to compute \(f(x) - 1\):

\[ f(x) - 1 = \frac{1 + 6x^2}{1 + 4x^2} - 1 = \frac{1 + 6x^2 - (1 + 4x^2)}{1 + 4x^2} = \frac{6x^2 - 4x^2}{1 + 4x^2} = \frac{2x^2}{1 + 4x^2}. \]

### Step 4: Substitute into the limit
Now we substitute \(f(x) - 1\) back into our limit:

\[ \lim_{x \to 0} \frac{1}{x^2} \cdot \left(\frac{2x^2}{1 + 4x^2}\right). \]

This simplifies to:

\[ \lim_{x \to 0} \frac{2x^2}{x^2(1 + 4x^2)} = \lim_{x \to 0} \frac{2}{1 + 4x^2}. \]

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

\[ \lim_{x \to 0} \frac{2}{1 + 4x^2} = \frac{2}{1 + 0} = 2. \]

### Step 6: Final result
Now we can find the original limit:

\[ L = e^{\lim_{x \to 0} g(x) \cdot (f(x) - 1)} = e^{2}. \]

Thus, the final result is:

\[ \lim_{x \to 0} \left(\frac{1 + 6x^2}{1 + 4x^2}\right)^{\frac{1}{x^2}} = e^{2}. \]