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Let’s check out your problem:

h(x)=(7-4x)/(3+sqrt(9x-2))
We want to find
lim_(x rarr3)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.

$h(x)=\frac{7-4 x}{3+\sqrt{9 x-2}}$ $\newline$ We want to find $\lim _{x \rightarrow 3} h(x)$ . $\newline$ 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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Full solution

Q.

h(x)=\frac{7-4 x}{3+\sqrt{9 x-2}}

\newline

We want to find

\lim _{x \rightarrow 3} 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.

Calculate $h(3)$ : Now, let's do the calculations. $\newline$ $h(3) = (7 - 12) / (3 + \sqrt{27 - 2})$ $\newline$ $h(3) = (-5) / (3 + \sqrt{25})$ $\newline$ $h(3) = (-5) / (3 + 5)$
Simplify expression: Simplify the expression. $h(3) = \frac{-5}{8}$
Final answer: Since we got a numerical value, the limit exists and we found it. $\newline$ So, the answer is (A) The limit exists, and we found it!

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