Question

\(\operatorname{Lt}_{x \rightarrow-2} \frac{x+2}{\sqrt{2 x+8}-\sqrt{2-x}}=\) (1) \(\frac{1}{3}\) (2) \(\frac{2}{3}\) (3) 1 (4) \(\frac{4}{3}\)

Short answer

Based on the solution above, the answer to the limit problem is: \(\operatorname{Lt}_{x \rightarrow -2} \frac{x+2}{\sqrt{2x+8} - \sqrt{2-x}} = 0\)

Step-by-step solution

Identify the given function and x-value

We are given the limit function: \(\operatorname{Lt}_{x \rightarrow -2} \frac{x+2}{\sqrt{2x+8} - \sqrt{2-x}}\) Our goal is to find the value of this limit as x approaches -2.

Simplify the expression

Before determining the limit, let us first analyze the expression and simplify the function if needed. We can see a sum of two square roots in the denominator. In such cases, we can use rationalization to find the limit. To do this, multiply the numerator and the denominator by the conjugate of the denominator: \(\sqrt{2x+8} + \sqrt{2-x}\) \[ \frac{x+2}{\sqrt{2x+8} - \sqrt{2-x}} \cdot \frac{\sqrt{2x+8}+\sqrt{2-x}}{\sqrt{2x+8}+\sqrt{2-x}} = \] \[ \frac{(x+2)(\sqrt{2x+8} + \sqrt{2-x})}{(\sqrt{2x+8} - \sqrt{2-x})(\sqrt{2x+8} + \sqrt{2-x})} \] Square the difference in the denominator \[ \frac{(x+2)(\sqrt{2x+8} + \sqrt{2-x})}{(2x+8-(2-x))} \] Simplify the denominator and we have: \[ \frac{(x+2)(\sqrt{2x+8} + \sqrt{2-x})}{2x+8-2+x} = \frac{(x+2)(\sqrt{2x+8} + \sqrt{2-x})}{3x+6} \]

Evaluate the limit

Now, let's plug in the value of x as -2. \[ \operatorname{Lt}_{x \rightarrow -2} \frac{(x+2)(\sqrt{2x+8} + \sqrt{2-x})}{3x+6} = \frac{(-2+2)(\sqrt{2(-2)+8} + \sqrt{2-(-2)})}{3(-2)+6} \] In this case, you can see plugging in the value results in the numerator being 0. Simplify and get the answer: \[ \frac{0}{-6+6} = 0 \] So, none of the given options is correct.

Key concepts

Rationalization Technique

The rationalization technique is a method used to eliminate radicals from the denominator or numerator of a fraction. In the context of limits, this is particularly useful when dealing with expressions involving square roots, which can cause indeterminate forms.

In our exercise, we had a complex fraction with square roots in the denominator: \[ \frac{x+2}{\sqrt{2x+8} - \sqrt{2-x}} \]To solve this, we multiplied both the numerator and the denominator by the conjugate of the denominator. The conjugate in this case is:
\[ \sqrt{2x+8} + \sqrt{2-x} \]This process helps to eliminate the square roots when multiplying the denominator by its conjugate, ultimately simplifying the expression. Remember, the product of a binomial and its conjugate results in a difference of squares. This simplification is key to allowing further steps without the complications of radicals.

Using the rationalization technique transforms what initially appears as a complex limit problem into a solvable one, eventually enabling us to substitute the limit value without encountering undefined expressions.

Square Root Simplification

Simplifying square roots is a fundamental skill in tackling problems with radicals, especially those involving limits. When square roots are in the denominator, as we encountered in this exercise, simplifying them ensures the expression is manageable and not undefined as you evaluate the limit.

In the solution, after using rationalization, the expression was simplified considerably. Let's break down the simplification of the square roots:
- The expression in the denominator, after rationalization, became:
\[ (\sqrt{2x+8} - \sqrt{2-x})(\sqrt{2x+8} + \sqrt{2-x}) = (2x+8) - (2-x) \]This simplification is crucial as it allows us to identify that the squares of the square roots cancel out, leading to a simpler polynomial expression.

Once simplified, substituting the approached value for \(x\) becomes straightforward, ensuring that square roots do not obscure the path to finding the limit. By practicing square root simplification, you build a toolkit that helps manage and resolve expressions that at first glance appear dauntingly complex.

Mathematical Problem-Solving

Mathematical problem-solving is the cornerstone of successfully navigating exercises involving calculus limits. This skill involves breaking down complex problems into simpler parts and systematically applying mathematical principles to arrive at a solution.

In our exercise, problem-solving began with correctly identifying that rationalization was necessary to deal with the square roots in the denominator. From there, we methodically proceeded to multiply by the conjugate, simplify where possible, and ultimately substitute the limit value.
- Problem-solving often benefits from:

• Recognizing patterns in expressions that could lead to simplification, such as differences of squares.
• Considering alternative strategies, like substitution or factoring, when confronted with difficult forms.
• Using logical reasoning to ascertain each step's validity, ensuring the approach aligns with mathematical laws.

This structured approach equips you to tackle new and unfamiliar problems, creating a reliable framework that encourages optimal results. Building your problem-solving capabilities enhances your ability to approach calculus limits and other mathematical puzzles confidently and effectively.