Find the limit of y = \(\frac{(x^3 - 2x^2 + 6x - 12)}{(x - 2)}\) as x goes to 2.

Answer Details

The limit of the function \( y = \frac{x^3 - 2x^2 + 6x - 12}{x - 2} \) as \( x \) approaches 2 can be found by first checking what happens when we substitute \( x = 2 \) directly.

Substituting \( x = 2 \):
Numerator: \( 2^3 - 2 \cdot 2^2 + 6 \cdot 2 - 12 = 8 - 8 + 12 - 12 = 0 \)
Denominator: \( 2 - 2 = 0 \)
So, the expression gives \( \frac{0}{0} \), which is an indeterminate form. To find the limit, we need to simplify the expression.

Let’s factor the numerator to see if \( (x - 2) \) is a factor:

x^3 - 2x^2 + 6x - 12

We can try to divide the numerator by \( (x - 2) \) using either polynomial division or synthetic division. Let's use synthetic division with a root of 2:

2 |  1   -2    6   -12
      2    0    12
   --------------------
      1    0    6     0

The result shows that \( x^3 - 2x^2 + 6x - 12 = (x - 2)(x^2 + 6) \).

So, we can rewrite the function:

\[ y = \frac{(x - 2)\big(x^2 + 6\big)}{x - 2} \]

As long as \( x \neq 2 \), the \( (x - 2) \) terms cancel:

\[ y = x^2 + 6 \]

Now, find the limit as \( x \to 2 \):

\[ \lim_{x \to 2} y = (2)^2 + 6 = 4 + 6 = 10 \]

Therefore, the limit as \( x \) approaches 2 is 10.

• The expression approaches 10 as \( x \) gets close to 2.
• It does NOT approach infinity or zero, and it does NOT approach 12.
• We found this by simplifying the original expression and evaluating the resulting function at \( x = 2 \).