f(x)=(x+2)/(x^(2)-3x+2) We want to find lim_(x rarr3)f(x). What happens when we use direct substitution? Choose 1 answer: (A) The limit exists, and we found it! (B) The limit doesn't exist (probably an asymptote). (C) The result is indeterminate.

Direct Substitution: Let's first try direct substitution by plugging x = 3 into the function f(x) to see what we get. f(3) = (3+2) / (3^2 - 3*3 + 2)Performing Calculations: Now, let's perform the calculations: f(3) = (5) / (9 - 9 + 2)Simplifying the Expression: Simplify the expression: f(3) = 5 / (0 + 2) f(3) = 5 / 2Existence of the Limit: Since we were able to find a value by direct substitution without encountering division by zero or any other undefined operations, the limit exists and we have found it.

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f(x)=(x+2)/(x^(2)-3x+2)
We want to find
lim_(x rarr3)f(x).
What happens when we use direct substitution?
Choose 1 answer:
(A) The limit exists, and we found it!
(B) The limit doesn't exist (probably an asymptote).
(C) The result is indeterminate.

$f(x)=\frac{x+2}{x^{2}-3 x+2}$ $\newline$ We want to find $\lim _{x \rightarrow 3} f(x)$ . $\newline$ What happens when we use direct substitution? $\newline$ Choose $1$ answer: $\newline$ (A) The limit exists, and we found it! $\newline$ (B) The limit doesn't exist (probably an asymptote). $\newline$ (C) The result is indeterminate.

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Full solution

f(x)=\frac{x+2}{x^{2}-3 x+2}

\newline

We want to find

\lim _{x \rightarrow 3} f(x)

\newline

What happens when we use direct substitution?

\newline

Choose

1

answer:

\newline

(A) The limit exists, and we found it!

\newline

(B) The limit doesn't exist (probably an asymptote).

\newline

Direct Substitution: Let's first try direct substitution by plugging $x = 3$ into the function $f(x)$ to see what we get. $\newline$ $f(3) = \frac{3+2}{3^2 - 3\times3 + 2}$
Performing Calculations: Now, let's perform the calculations: $f(3) = \frac{5}{9 - 9 + 2}$
Simplifying the Expression: Simplify the expression: $\newline$ $f(3) = \frac{5}{(0 + 2)}$ $\newline$ $f(3) = \frac{5}{2}$
Existence of the Limit: Since we were able to find a value by direct substitution without encountering division by $0$ or any other undefined operations, the limit exists and we have found it.