You uploaded an image containing the following limit expression:

$\lim_{x \to 1} \frac{x^2 - 1}{\sqrt{x + 3} - \sqrt{4x}}$

Let's solve it step by step.

Step 1: Factorize the numerator

The numerator $x^2 - 1$ is a difference of squares, so it can be factored as:

$x^2 - 1 = (x - 1)(x + 1)$

Thus, the expression becomes:

$\frac{(x - 1)(x + 1)}{\sqrt{x + 3} - \sqrt{4x}}$

Step 2: Rationalize the denominator

To deal with the square roots in the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{x + 3} + \sqrt{4x}$:

$\frac{(x - 1)(x + 1)}{(\sqrt{x + 3} - \sqrt{4x})} \times \frac{\sqrt{x + 3} + \sqrt{4x}}{\sqrt{x + 3} + \sqrt{4x}}$

This gives:

$\frac{(x - 1)(x + 1)(\sqrt{x + 3} + \sqrt{4x})}{(\sqrt{x + 3})^2 - (\sqrt{4x})^2}$

Step 3: Simplify the denominator

Simplifying the denominator using the difference of squares:

$(\sqrt{x + 3})^2 - (\sqrt{4x})^2 = (x + 3) - (4x) = -3x + 3 = 3(1 - x)$

Thus, the expression simplifies to:

$\frac{(x - 1)(x + 1)(\sqrt{x + 3} + \sqrt{4x})}{3(1 - x)}$

Step 4: Cancel common terms

We can cancel $x - 1$ with $1 - x$, noting that $x - 1 = -(1 - x)$. So the expression becomes:

$\frac{-(x + 1)(\sqrt{x + 3} + \sqrt{4x})}{3}$

Step 5: Evaluate the limit

Now, substitute $x = 1$ into the expression:

$\frac{-(1 + 1)(\sqrt{1 + 3} + \sqrt{4 \times 1})}{3} = \frac{-2(\sqrt{4} + \sqrt{4})}{3} = \frac{-2(2 + 2)}{3} = \frac{-8}{3}$

Thus, the value of the limit is:

$\boxed{-\frac{8}{3}}$

Would you like more details on any part of this solution?

5 Relative Questions:

1. How is rationalizing the denominator useful in solving limits with square roots?
2. What other types of factoring can be used to simplify limits like this?
3. How do conjugates help simplify expressions involving square roots?
4. Why do we cancel terms before substituting the limit value?
5. What is the significance of the difference of squares in calculus?

Tip:

When dealing with limits involving square roots, rationalizing the denominator can greatly simplify the expression, making it easier to evaluate the limit.