The limit problem shown in the image is:

$\lim_{{x \to 0}} \frac{\sqrt{x+1} - 1}{x}$

To solve this type of limit, where direct substitution results in an indeterminate form $\frac{0}{0}$, you can use algebraic manipulation such as multiplying by the conjugate of the numerator. Here's the step-by-step approach:

Step 1: Multiply by the conjugate

Multiply both the numerator and the denominator by the conjugate of the numerator, which is $\sqrt{x+1} + 1$:

$\frac{\sqrt{x+1} - 1}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1} = \frac{(\sqrt{x+1} - 1)(\sqrt{x+1} + 1)}{x (\sqrt{x+1} + 1)}$

Step 2: Simplify the numerator

Using the difference of squares formula $(a - b)(a + b) = a^2 - b^2$, simplify the numerator:

$(\sqrt{x+1})^2 - 1^2 = (x+1) - 1 = x$

Thus, the expression becomes:

$\frac{x}{x(\sqrt{x+1} + 1)} = \frac{1}{\sqrt{x+1} + 1}$

Step 3: Apply the limit

Now, substitute $x = 0$ directly into the simplified expression:

$\lim_{{x \to 0}} \frac{1}{\sqrt{x+1} + 1} = \frac{1}{\sqrt{0+1} + 1} = \frac{1}{1+1} = \frac{1}{2}$

Final Answer:

$\lim_{{x \to 0}} \frac{\sqrt{x+1} - 1}{x} = \frac{1}{2}$

Would you like more details on any of these steps, or do you have any questions?

Related Questions:

1. How do you recognize indeterminate forms in limit problems?
2. What are other common techniques for solving limits, aside from multiplying by the conjugate?
3. How do you handle limits involving complex fractions?
4. What is L'Hopital's Rule, and when should it be used?
5. Can this type of problem also be solved using a series expansion?

Tip:

Always check if the direct substitution yields an indeterminate form. If it does, think about algebraic simplifications like multiplying by conjugates, factoring, or applying L'Hopital's Rule.