Evaluate the following limits :
`Lim_( xto 0) (sqrt(1+x) -sqrt(1-x))/(2x)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{2x}, \]

we can follow these steps:

### Step 1: Multiply by the Conjugate
We will multiply the numerator and the denominator by the conjugate of the numerator, which is \(\sqrt{1+x} + \sqrt{1-x}\).

\[ \lim_{x \to 0} \frac{(\sqrt{1+x} - \sqrt{1-x})(\sqrt{1+x} + \sqrt{1-x})}{2x(\sqrt{1+x} + \sqrt{1-x})} \]

### Step 2: Simplify the Numerator
Using the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\), we can simplify the numerator:

\[ \sqrt{1+x}^2 - \sqrt{1-x}^2 = (1+x) - (1-x) = x + x = 2x. \]

Thus, we have:

\[ \lim_{x \to 0} \frac{2x}{2x(\sqrt{1+x} + \sqrt{1-x})}. \]

### Step 3: Cancel Out Common Factors
Now we can cancel \(2x\) in the numerator and denominator:

\[ \lim_{x \to 0} \frac{1}{\sqrt{1+x} + \sqrt{1-x}}. \]

### Step 4: Substitute \(x = 0\)
Now, we can substitute \(x = 0\) into the limit:

\[ \frac{1}{\sqrt{1+0} + \sqrt{1-0}} = \frac{1}{\sqrt{1} + \sqrt{1}} = \frac{1}{1 + 1} = \frac{1}{2}. \]

### Final Answer
Thus, the limit is

\[ \frac{1}{2}. \]