Limit Calculation for Sum of Functions

The problem asks for the limit of the sum of two functions, $f_1(x)$ and $f_2(x)$, as $x$ approaches 2.

Given functions:

• $f_1(x) = \frac{x^2 - 4}{x - 2}$
• $f_2(x) = x^2 - 2x + 2$

Evaluating $f_1(x)$ as $x \to 2$

The function $f_1(x)$ is undefined at $x=2$ (indeterminate form $\frac{0}{0}$). We simplify it first:

$f_1(x) = \frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2}$

For $x \neq 2$, we can cancel the $(x - 2)$ terms:

$f_1(x) = x + 2$

Now, we find the limit:

$\lim_{x \to 2} f_1(x) = \lim_{x \to 2} (x + 2) = 2 + 2 = 4$

Evaluating $f_2(x)$ as $x \to 2$

$f_2(x)$ is a polynomial function, so we can directly substitute $x=2$ to find its limit:

$\lim_{x \to 2} f_2(x) = \lim_{x \to 2} (x^2 - 2x + 2) = (2)^2 - 2(2) + 2$

$= 4 - 4 + 2 = 2$

Finding the Limit of the Sum $(f_1(x) + f_2(x))$

The limit of a sum is the sum of the limits, provided the individual limits exist:

$\lim_{x \to 2} (f_1(x) + f_2(x)) = \lim_{x \to 2} f_1(x) + \lim_{x \to 2} f_2(x)$

Substituting the calculated limits:

$= 4 + 2 = 6$

Conclusion

The value of $(f_1(x) + f_2(x))$ as $x \to 2$ is 6.