$lim_{x\rightarrow2}[\frac{x^{2}-4}{x^{3}-4x^{2}+4x}]=$

Q: $lim_{x\rightarrow2}[\frac{x^{2}-4}{x^{3}-4x^{2}+4x}]=$

To solve the limit problem $ \lim_{x \rightarrow 2} \left[\frac{x^{2}-4}{x^{3}-4x^{2}+4x}\right] $, we need to analyze the behavior of the function as $ x $ approaches 2. First, let's simplify the expression: 1. **Numerator**: $ x^2 - 4 $ can be factored as $ (x-2)(x+2) $. 2. **Denominator**: $ x^3 - 4x^2 + 4x $ can be factored by first factoring out an $ x $, giving $ x(x^2 - 4x + 4) $. The quadratic $ x^2 - 4x + 4 $ can be factored further as $ (x-2)^2 $. Thus, the expression becomes: \[ \frac{(x-2)(x+2)}{x(x-2)^2} \] Now, cancel the common factor $ (x-2) $ from the numerator and the denominator: \[ \frac{x+2}{x(x-2)} \] Now, evaluate the limit as $ x \rightarrow 2 $: \[ \lim_{x \rightarrow 2} \frac{x+2}{x(x-2)} \] As $ x $ approaches 2, the denominator $ x(x-2) $ approaches 0 because $ (x-2) $ approaches 0, while the numerator $ x+2 $ approaches 4. This results in a form where the numerator is a constant (4) and the denominator approaches 0, which indicates that the limit approaches infinity. Therefore, the correct answer is Option 4: $ \infty $. **Explanation for Incorrect Options:** 1. **Option 1 (0)**: This option is incorrect because the numerator does not approach 0 as $ x \rightarrow 2 $. Instead, it approaches 4, while the denominator approaches 0, leading to an infinite limit. 2. **Option 2 (2)**: This option is incorrect because the limit does not simplify to a finite number like 2. The behavior of the function as $ x \rightarrow 2 $ leads to an infinite result, not a finite one. 3. **Option 3 (4)**: This option is incorrect because, although the numerator approaches 4, the denominator approaches 0, which results in an infinite limit, not a finite value like 4. In conclusion, the limit evaluates to infinity as $ x $ approaches 2, supporting Option 4 as the correct answer.

Question

Accepted Answer

To solve the limit problem $ \lim_{x \rightarrow 2} \left[\frac{x^{2}-4}{x^{3}-4x^{2}+4x}\right] $, we need to analyze the behavior of the function as $ x $ approaches 2.

First, let's simplify the expression:

1. **Numerator**: $ x^2 - 4 $ can be factored as $ (x-2)(x+2) $.

2. **Denominator**: $ x^3 - 4x^2 + 4x $ can be factored by first factoring out an $ x $, giving $ x(x^2 - 4x + 4) $. The quadratic $ x^2 - 4x + 4 $ can be factored further as $ (x-2)^2 $.

Thus, the expression becomes:
$$ \frac{(x-2)(x+2)}{x(x-2)^2} $$

Now, cancel the common factor $ (x-2) $ from the numerator and the denominator:
$$ \frac{x+2}{x(x-2)} $$

Now, evaluate the limit as $ x \rightarrow 2 $:
$$ \lim_{x \rightarrow 2} \frac{x+2}{x(x-2)} $$

As $ x $ approaches 2, the denominator $ x(x-2) $ approaches 0 because $ (x-2) $ approaches 0, while the numerator $ x+2 $ approaches 4. This results in a form where the numerator is a constant (4) and the denominator approaches 0, which indicates that the limit approaches infinity.

Therefore, the correct answer is Option 4: $ \infty $.

**Explanation for Incorrect Options:**

1. **Option 1 (0)**: This option is incorrect because the numerator does not approach 0 as $ x \rightarrow 2 $. Instead, it approaches 4, while the denominator approaches 0, leading to an infinite limit.

2. **Option 2 (2)**: This option is incorrect because the limit does not simplify to a finite number like 2. The behavior of the function as $ x \rightarrow 2 $ leads to an infinite result, not a finite one.

3. **Option 3 (4)**: This option is incorrect because, although the numerator approaches 4, the denominator approaches 0, which results in an infinite limit, not a finite value like 4.

In conclusion, the limit evaluates to infinity as $ x $ approaches 2, supporting Option 4 as the correct answer.