Math Problem Statement

\lim _{x\to :2}\left(\frac{\sqrt{x-2}+x^2-3x+2}{\sqrt{x^2-4}}\right)

Solution

We are tasked with evaluating the limit:

limx2(x2+x23x+2x24)\lim_{x \to 2} \left( \frac{\sqrt{x - 2} + x^2 - 3x + 2}{\sqrt{x^2 - 4}} \right)

Step 1: Simplify the expression

First, let's analyze the denominator and numerator separately.

Denominator:

x24\sqrt{x^2 - 4} This can be factored as: x24=(x2)(x+2)\sqrt{x^2 - 4} = \sqrt{(x - 2)(x + 2)}

Numerator:

x2+x23x+2\sqrt{x - 2} + x^2 - 3x + 2 Now, simplify the quadratic part x23x+2x^2 - 3x + 2: x23x+2=(x1)(x2)x^2 - 3x + 2 = (x - 1)(x - 2) So the numerator becomes: x2+(x1)(x2)\sqrt{x - 2} + (x - 1)(x - 2)

Thus, the entire expression now is: limx2x2+(x1)(x2)(x2)(x+2)\lim_{x \to 2} \frac{\sqrt{x - 2} + (x - 1)(x - 2)}{\sqrt{(x - 2)(x + 2)}}

Step 2: Apply the limit as x2x \to 2

When x2x \to 2, both the numerator and denominator contain the term x2\sqrt{x - 2}, which tends to zero. Hence, we have an indeterminate form of type 00\frac{0}{0}, so we need to simplify further or apply L'Hôpital's Rule.

Step 3: Factor and Cancel

Factor (x2)(x - 2) from the numerator and denominator: limx2(x2)(1x2+(x1))(x2)(x+2)\lim_{x \to 2} \frac{(x - 2)(\frac{1}{\sqrt{x - 2}} + (x - 1))}{\sqrt{(x - 2)(x + 2)}}

We can cancel x2\sqrt{x - 2} from both the numerator and denominator: limx21x2+(x1)x+2\lim_{x \to 2} \frac{\frac{1}{\sqrt{x - 2}} + (x - 1)}{\sqrt{x + 2}}

Step 4: Apply the limit

Now, substitute x=2x = 2 directly: 122+(21)=undefined because the first term involves division by zero.\frac{1}{\sqrt{2 - 2}} + (2 - 1) = \text{undefined because the first term involves division by zero}.

Thus, the limit does not exist because the expression blows up as x2x \to 2.

The final result is:

The limit does not exist.\boxed{\text{The limit does not exist.}}

Would you like further explanation or details?


Here are 5 related questions:

  1. How does L'Hôpital's Rule help resolve indeterminate forms?
  2. What are common methods for factoring expressions involving square roots?
  3. What is the difference between limits that exist and those that do not?
  4. How do you approach limits where square roots appear in both the numerator and denominator?
  5. Can a limit approaching infinity ever result in a defined value?

Tip: Always try substituting directly before applying advanced techniques like L'Hôpital's Rule or factoring.

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Math Problem Analysis

Mathematical Concepts

Limits
Indeterminate Forms
Square Roots
Factoring

Formulas

L'Hôpital's Rule
Square Root Simplification
Quadratic Factoring

Theorems

L'Hôpital's Rule
Limit Definition

Suitable Grade Level

Undergraduate Calculus