h(x)=(sqrt(54+x)-7)/(6x+30) We want to find lim_(x rarr-5)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.

Direct Substitution: First, let's try direct substitution of x = -5 into the function h(x) to see what we get. h(x) = (sqrt(54 + x) - 7) / (6x + 30) h(-5) = (sqrt(54 - 5) - 7) / (6(-5) + 30)Perform Calculations: Now, let's perform the calculations inside the square root and the denominator. h(-5) = (sqrt(49) - 7) / ( -30 + 30)Simplify Square Root and Denominator: Simplify the square root and the denominator. h(-5) = (7 - 7) / 0Indeterminate Form: We see that the numerator simplifies to 0, but the denominator also simplifies to 0. h(-5) = 0 / 0 This is an indeterminate form, which means we cannot determine the limit through direct substitution.

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h(x)=(sqrt(54+x)-7)/(6x+30)
We want to find
lim_(x rarr-5)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.

$h(x)=\frac{\sqrt{54+x}-7}{6 x+30}$ $\newline$ We want to find $\lim _{x \rightarrow-5} h(x)$ . $\newline$ 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

h(x)=\frac{\sqrt{54+x}-7}{6 x+30}

\newline

We want to find

\lim _{x \rightarrow-5} 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

Direct Substitution: First, let's try direct substitution of $x = -5$ into the function $h(x)$ to see what we get. $\newline$ $h(x) = \frac{\sqrt{54 + x} - 7}{6x + 30}$ $\newline$ $h(-5) = \frac{\sqrt{54 - 5} - 7}{6(-5) + 30}$
Perform Calculations: Now, let's perform the calculations inside the square root and the denominator. $\newline$ $h(-5) = (\sqrt{49} - 7) / ( -30 + 30)$
Simplify Square Root and Denominator: Simplify the square root and the denominator. $h(-5) = \frac{7 - 7}{0}$
Indeterminate Form: We see that the numerator simplifies to $0$ , but the denominator also simplifies to $0$ . $h(-5) = \frac{0}{0}$ This is an indeterminate form, which means we cannot determine the limit through direct substitution.