Evaluate the following limits :
`lim_(x to 0)(sqrt(1+x)-sqrt(1-x))/x`

Answer

To evaluate the limit

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

we can follow these steps:

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

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

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

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

So now we have:

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

### Step 3: Simplify the Expression
We can cancel \(x\) in the numerator and denominator (as long as \(x \neq 0\)):

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

### Step 4: Substitute \(x = 0\)
Now we can substitute \(x = 0\):

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

### Final Answer
Thus, the value of the limit is

\[ \boxed{1}. \]