h(x)=(3x-11)/(x^(2)-1) We want to find lim_(x rarr1)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: Let's first attempt to use direct substitution to find the limit as x approaches 1 for the function h(x)=(3x-11)/(x^2-1). Substitute x = 1 into the function: h(1) = (3(1) - 11) / ((1)^2 - 1) h(1) = (3 - 11) / (1 - 1) h(1) = (-8) / (0)Substituting x = 1: We observe that the denominator becomes 0, which means the function is undefined at x = 1. This results in a division by zero, which is undefined in mathematics. Since we have a non-zero numerator and a zero denominator, the result is an indeterminate form.Undefined at x = 1: The indeterminate form indicates that we cannot determine the limit by direct substitution. We need to use other methods to evaluate the limit, such as factoring, simplifying, or applying L'Hôpital's rule if necessary. However, for the purpose of answering the multiple-choice question, we have enough information to conclude that direct substitution leads to an indeterminate form.

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h(x)=(3x-11)/(x^(2)-1)
We want to find
lim_(x rarr1)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.

$h(x)=\frac{3 x-11}{x^{2}-1}$ $\newline$ We want to find $\lim _{x \rightarrow 1} h(x)$ . $\newline$ 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{3 x-11}{x^{2}-1}

\newline

We want to find

\lim _{x \rightarrow 1} 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: Let's first attempt to use direct substitution to find the limit as $x$ approaches $1$ for the function $h(x)=\frac{3x-11}{x^2-1}$ . Substitute $x = 1$ into the function: $h(1) = \frac{3(1) - 11}{(1)^2 - 1}$ $h(1) = \frac{3 - 11}{1 - 1}$ $h(1) = \frac{-8}{0}$
Substituting $x = 1$ : We observe that the denominator becomes $0$ , which means the function is undefined at $x = 1$ . This results in a division by zero, which is undefined in mathematics. $\newline$ Since we have a non-zero numerator and a zero denominator, the result is an indeterminate form.
Undefined at $x = 1$ : The indeterminate form indicates that we cannot determine the limit by direct substitution. We need to use other methods to evaluate the limit, such as factoring, simplifying, or applying L'Hôpital's rule if necessary. $\newline$ However, for the purpose of answering the multiple-choice question, we have enough information to conclude that direct substitution leads to an indeterminate form.