lim_(x rarr0)csc(x)=? Choose 1 answer: (A) -1 (B) 0 (c) 1 (D) The limit doesn't exist.

Define csc(x): We know that csc(x) is the reciprocal of sin(x), so csc(x) = 1/sin(x). To find the limit of csc(x) as x approaches 0, we need to consider the behavior of sin(x) near 0.Consider sin(x) behavior: Since sin(0) = 0, the reciprocal 1/sin(x) will become very large as x approaches 0. This means that the limit of csc(x) as x approaches 0 is not finite.Reciprocal becomes large: We must also consider the behavior from both sides of 0. As x approaches 0 from the positive side, sin(x) is positive, and thus csc(x) is positive and grows without bound. As x approaches 0 from the negative side, sin(x) is negative, and thus csc(x) is negative and decreases without bound.Behavior from both sides: Since the behavior of csc(x) as x approaches 0 from the positive and negative sides is not consistent (it does not approach a single finite number), the limit does not exist.

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lim_(x rarr0)csc(x)=?
Choose 1 answer:
(A) -1
(B) 0
(c) 1
(D) The limit doesn't exist.

$\lim _{x \rightarrow 0} \csc (x)=?$ $\newline$ Choose $1$ answer: $\newline$ (A) $-1$ $\newline$ (B) $0$ $\newline$ (C) $1$ $\newline$ (D) The limit doesn't exist.

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Algebra 2

Inverses of sin, cos, and tan: degrees

Full solution

\lim _{x \rightarrow 0} \csc (x)=?

\newline

Choose

1

answer:

\newline

(A)

-1

\newline

(B)

0

\newline

(C)

1

\newline

(D) The limit doesn't exist.

Define $csc(x)$ : We know that $csc(x)$ is the reciprocal of $\sin(x)$ , so $csc(x) = \frac{1}{\sin(x)}$ . To find the limit of $csc(x)$ as $x$ approaches $0$ , we need to consider the behavior of $\sin(x)$ near $0$ .
Consider $\sin(x)$ behavior: Since $\sin(0) = 0$ , the reciprocal $\frac{1}{\sin(x)}$ will become very large as $x$ approaches $0$ . This means that the limit of $\csc(x)$ as $x$ approaches $0$ is not finite.
Reciprocal becomes large: We must also consider the behavior from both sides of $0$ . As $x$ approaches $0$ from the positive side, $\sin(x)$ is positive, and thus $\csc(x)$ is positive and grows without bound. As $x$ approaches $0$ from the negative side, $\sin(x)$ is negative, and thus $\csc(x)$ is negative and decreases without bound.
Behavior from both sides: Since the behavior of $\csc(x)$ as $x$ approaches $0$ from the positive and negative sides is not consistent (it does not approach a single finite number), the limit does not exist.