Evaluate the following limits: `lim_(xrarr0)((sqrt(1+x^(2))-sqrt(1+x))/(sqrt(1+x^(3))-sqrt(1+x)))`

Explanation

To evaluate the limit

\[ \lim_{x \to 0} \frac{\sqrt{1+x^2} - \sqrt{1+x}}{\sqrt{1+x^3} - \sqrt{1+x}}, \]

we will follow these steps:

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

\[ \frac{\sqrt{1+0^2} - \sqrt{1+0}}{\sqrt{1+0^3} - \sqrt{1+0}} = \frac{\sqrt{1} - \sqrt{1}}{\sqrt{1} - \sqrt{1}} = \frac{0}{0}. \]

This is an indeterminate form, so we need to manipulate the expression.

**Hint:** Check if the limit results in an indeterminate form before proceeding.

### Step 2: Multiply by the conjugate
To resolve the indeterminate form, we can multiply the numerator and denominator by the conjugate of both the numerator and the denominator.

The conjugate of the numerator \( \sqrt{1+x^2} - \sqrt{1+x} \) is \( \sqrt{1+x^2} + \sqrt{1+x} \), and the conjugate of the denominator \( \sqrt{1+x^3} - \sqrt{1+x} \) is \( \sqrt{1+x^3} + \sqrt{1+x} \).

Thus, we rewrite the limit as:

\[ \lim_{x \to 0} \frac{(\sqrt{1+x^2} - \sqrt{1+x})(\sqrt{1+x^2} + \sqrt{1+x})}{(\sqrt{1+x^3} - \sqrt{1+x})(\sqrt{1+x^3} + \sqrt{1+x})}. \]

**Hint:** Multiplying by the conjugate helps eliminate the square roots.

### Step 3: Simplify the expression
Using the difference of squares, we can simplify the numerator and denominator:

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

- Denominator:
\[ (\sqrt{1+x^3})^2 - (\sqrt{1+x})^2 = (1+x^3) - (1+x) = x^3 - x. \]

Now, we can rewrite the limit:

\[ \lim_{x \to 0} \frac{x^2 - x}{x^3 - x} \cdot \frac{\sqrt{1+x^2} + \sqrt{1+x}}{\sqrt{1+x^3} + \sqrt{1+x}}. \]

**Hint:** Use the difference of squares to simplify the expressions.

### Step 4: Factor the expressions
We can factor out \( x \) from both the numerator and the denominator:

\[ x(x - 1) \text{ in the numerator and } x(x^2 - 1) \text{ in the denominator}. \]

Thus, we have:

\[ \lim_{x \to 0} \frac{x(x - 1)}{x(x^2 - 1)} \cdot \frac{\sqrt{1+x^2} + \sqrt{1+x}}{\sqrt{1+x^3} + \sqrt{1+x}}. \]

Cancelling \( x \) (since \( x \neq 0 \) in the limit),

\[ \lim_{x \to 0} \frac{x - 1}{x^2 - 1} \cdot \frac{\sqrt{1+x^2} + \sqrt{1+x}}{\sqrt{1+x^3} + \sqrt{1+x}}. \]

**Hint:** Factor out common terms to simplify further.

### Step 5: Evaluate the limit
Now we can substitute \( x = 0 \):

- The first part becomes:
\[ \frac{0 - 1}{0^2 - 1} = \frac{-1}{-1} = 1. \]

- The second part becomes:
\[ \frac{\sqrt{1+0^2} + \sqrt{1+0}}{\sqrt{1+0^3} + \sqrt{1+0}} = \frac{\sqrt{1} + \sqrt{1}}{\sqrt{1} + \sqrt{1}} = \frac{1 + 1}{1 + 1} = \frac{2}{2} = 1. \]

Thus, the limit evaluates to:

\[ 1 \cdot 1 = 1. \]

### Final Answer
The limit is

\[ \boxed{1}. \]