`lim_(xto0) (ln(2+x^(2))-ln(2-x^(2)))/(x^(2))` is equal to

Explanation

To solve the limit \( \lim_{x \to 0} \frac{\ln(2 + x^2) - \ln(2 - x^2)}{x^2} \), we can follow these steps:

### Step 1: Rewrite the limit using properties of logarithms
Using the property of logarithms that states \( \ln a - \ln b = \ln \frac{a}{b} \), we can rewrite the expression:
\[ \lim_{x \to 0} \frac{\ln\left(\frac{2 + x^2}{2 - x^2}\right)}{x^2} \]

### Step 2: Substitute \( x^2 \) with \( 2t \)
To simplify the limit, we can make the substitution \( x^2 = 2t \). As \( x \to 0 \), \( t \to 0 \) as well. Thus, we rewrite the limit:
\[ \lim_{t \to 0} \frac{\ln\left(\frac{2 + 2t}{2 - 2t}\right)}{2t} \]

### Step 3: Simplify the logarithmic expression
Now, we can simplify the expression inside the logarithm:
\[ \frac{2 + 2t}{2 - 2t} = \frac{2(1 + t)}{2(1 - t)} = \frac{1 + t}{1 - t} \]
Thus, our limit becomes:
\[ \lim_{t \to 0} \frac{\ln\left(\frac{1 + t}{1 - t}\right)}{2t} \]

### Step 4: Apply the logarithmic limit property
Using the property of logarithms, we can separate the logarithm:
\[ \ln\left(\frac{1 + t}{1 - t}\right) = \ln(1 + t) - \ln(1 - t) \]
So, our limit can be rewritten as:
\[ \lim_{t \to 0} \frac{\ln(1 + t) - \ln(1 - t)}{2t} \]

### Step 5: Use the standard limit
We know from calculus that:
\[ \lim_{u \to 0} \frac{\ln(1 + u)}{u} = 1 \]
Thus, we can apply this limit:
\[ \lim_{t \to 0} \frac{\ln(1 + t)}{t} = 1 \quad \text{and} \quad \lim_{t \to 0} \frac{\ln(1 - t)}{-t} = 1 \]
Therefore, we can express our limit as:
\[ \lim_{t \to 0} \frac{\ln(1 + t) - \ln(1 - t)}{2t} = \frac{1 + 1}{2} = 1 \]

### Final Result
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\ln(2 + x^2) - \ln(2 - x^2)}{x^2} = 1 \]