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

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} \), we can follow these steps:

### Step 1: Identify the limit
We start with the limit expression:
\[ \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} \]

### Step 2: Multiply by the conjugate
To simplify the expression, we can multiply the numerator and the denominator by the conjugate of the numerator, which is \( \sqrt{1+x} + 1 \):
\[ \lim_{x \to 0} \frac{(\sqrt{1+x} - 1)(\sqrt{1+x} + 1)}{x(\sqrt{1+x} + 1)} \]

### Step 3: Apply the difference of squares
Using the difference of squares formula \( a^2 - b^2 = (a-b)(a+b) \), we can simplify the numerator:
\[ \sqrt{1+x} - 1 = \frac{(1+x) - 1^2}{\sqrt{1+x} + 1} = \frac{x}{\sqrt{1+x} + 1} \]
Thus, we rewrite the limit:
\[ \lim_{x \to 0} \frac{x}{x(\sqrt{1+x} + 1)} = \lim_{x \to 0} \frac{1}{\sqrt{1+x} + 1} \]

### Step 4: Cancel the \( x \) terms
The \( x \) in the numerator and denominator cancels out:
\[ \lim_{x \to 0} \frac{1}{\sqrt{1+x} + 1} \]

### Step 5: Substitute \( x = 0 \)
Now we can substitute \( x = 0 \) into the limit:
\[ \frac{1}{\sqrt{1+0} + 1} = \frac{1}{1 + 1} = \frac{1}{2} \]

### Final Answer
Thus, the limit evaluates to:
\[ \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} = \frac{1}{2} \]

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