lim (square root of 5+2) -(square root of 5) divided by y   x approaches 0

It looks like you want to compute the following limit:

\[
\lim_{x \to 0} \frac{\sqrt{5 + 2x} - \sqrt{5}}{x}
\]

This is a typical indeterminate form of the type \( \frac{0}{0} \), so we can apply algebraic manipulation, specifically multiplying by the conjugate of the numerator to simplify.

### Step 1: Multiply by the conjugate
To get rid of the square roots in the numerator, multiply both the numerator and the denominator by the conjugate of \( \sqrt{5 + 2x} - \sqrt{5} \), which is \( \sqrt{5 + 2x} + \sqrt{5} \).

\[
\frac{\sqrt{5 + 2x} - \sqrt{5}}{x} \times \frac{\sqrt{5 + 2x} + \sqrt{5}}{\sqrt{5 + 2x} + \sqrt{5}}
\]

This gives us:

\[
\frac{(\sqrt{5 + 2x} - \sqrt{5})(\sqrt{5 + 2x} + \sqrt{5})}{x(\sqrt{5 + 2x} + \sqrt{5})}
\]

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

\[
(\sqrt{5 + 2x})^2 - (\sqrt{5})^2 = (5 + 2x) - 5 = 2x
\]

So now, the expression becomes:

\[
\frac{2x}{x(\sqrt{5 + 2x} + \sqrt{5})}
\]

### Step 3: Cancel the \(x\) terms
We can cancel \(x\) from the numerator and denominator:

\[
\frac{2}{\sqrt{5 + 2x} + \sqrt{5}}
\]

### Step 4: Take the limit as \(x \to 0\)
Now, substitute \(x = 0\) into the simplified expression:

\[
\frac{2}{\sqrt{5 + 2(0)} + \sqrt{5}} = \frac{2}{\sqrt{5} + \sqrt{5}} = \frac{2}{2\sqrt{5}} = \frac{1}{\sqrt{5}}
\]

So, the limit is:

\[
\lim_{x \to 0} \frac{\sqrt{5 + 2x} - \sqrt{5}}{x} = \frac{1}{\sqrt{5}}
\]

### Conclusion:
The limit evaluates to \( \frac{1}{\sqrt{5}} \).

Would you like further clarification or a breakdown of any steps? Here are some related questions you might find interesting:

1. How do limits involving square roots and indeterminate forms work in general?
2. What is the difference between applying the conjugate in algebraic limits vs. using L'Hopital's rule?
3. How do you deal with higher-order polynomials in limits like this?
4. Can you apply the same technique to more complex expressions, such as those involving trigonometric functions?
5. What happens when you try to compute the limit as \( x \to 0 \) for a function with an absolute value?

**Tip**: When dealing with square roots in limits, always check if multiplying by the conjugate can simplify the expression—this is especially helpful when you encounter indeterminate forms.

lim (square root of 5+2) - (square root of 5) divided by y x approaches 0

Learn how to compute the limit of the expression (sqrt(5+2x) - sqrt(5))/x as x approaches 0 using the conjugate method. This step-by-step guide covers indeterminate forms, limits, and simplifying algebraic expressions. Suitable for high school students in Grades 11-12.

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