h(x)=(x^(2)-6x+10)/(x-1) We want to find lim_(x rarr2)h(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.

Step 1: Direct Substitution: First, let's try direct substitution by plugging x = 2 into the function h(x). h(2) = ((2^2) - 6*2 + 10) / (2 - 1)Step 2: Calculation: Now, let's perform the calculations. h(2) = ((4 - 12 + 10) / (2 - 1)) h(2) = (2 / 1) h(2) = 2Step 3: Simplification: Since we were able to find a value by direct substitution without encountering division by zero or an indeterminate form, the limit exists. The limit as x approaches 2 for the function h(x) is 2.

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h(x)=(x^(2)-6x+10)/(x-1)
We want to find
lim_(x rarr2)h(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.

$h(x)=\frac{x^{2}-6 x+10}{x-1}$ $\newline$ We want to find $\lim _{x \rightarrow 2} h(x)$ . $\newline$ 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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Math Problems

Algebra 1

Write and solve inverse variation equations

Full solution

h(x)=\frac{x^{2}-6 x+10}{x-1}

\newline

We want to find

\lim _{x \rightarrow 2} h(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

Step $1$ : Direct Substitution: First, let's try direct substitution by plugging $x = 2$ into the function $h(x)$ . $h(2) = \frac{(2^2) - 6 \cdot 2 + 10}{2 - 1}$
Step $2$ : Calculation: Now, let's perform the calculations. $\newline$ $h(2) = \frac{(4 - 12 + 10)}{(2 - 1)}$ $\newline$ $h(2) = \frac{2}{1}$ $\newline$ $h(2) = 2$
Step $3$ : Simplification: Since we were able to find a value by direct substitution without encountering division by zero or an indeterminate form, the limit exists. $\newline$ The limit as $x$ approaches $2$ for the function $h(x)$ is $2$ .