g(x)=(tan(x))/(1-cos(x)) We want to find lim_(x rarr pi)g(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 try direct substitution of x = pi into the function g(x) = (tan(x))/(1 - cos(x)). g(pi) = (tan(pi))/(1 - cos(pi))Calculate Values: Calculate tan(pi) and cos(pi). tan(pi) = 0 (since the tangent of pi is 0) cos(pi) = -1 (since the cosine of pi is -1)Substitute Values: Substitute these values into the function. g(pi) = 0/(1 - (-1)) g(pi) = 0/(1 + 1) g(pi) = 0/2 g(pi) = 0Calculate Limit: Since we get a defined value (0) when we substitute x = pi, the limit exists and we have found it.

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g(x)=(tan(x))/(1-cos(x))
We want to find
lim_(x rarr pi)g(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.

$g(x)=\frac{\tan (x)}{1-\cos (x)}$ $\newline$ We want to find $\lim _{x \rightarrow \pi} g(x)$ . $\newline$ 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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g(x)=\frac{\tan (x)}{1-\cos (x)}

\newline

We want to find

\lim _{x \rightarrow \pi} g(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 try direct substitution of $x = \pi$ into the function $g(x) = \frac{\tan(x)}{1 - \cos(x)}$ . $g(\pi) = \frac{\tan(\pi)}{1 - \cos(\pi)}$
Calculate Values: Calculate $\tan(\pi)$ and $\cos(\pi)$ . $\newline$ $\tan(\pi) = 0$ (since the tangent of $\pi$ is $0$ ) $\newline$ $\cos(\pi) = -1$ (since the cosine of $\pi$ is $-1$ )
Substitute Values: Substitute these values into the function. $\newline$ $g(\pi) = \frac{0}{1 - (-1)}$ $\newline$ $g(\pi) = \frac{0}{1 + 1}$ $\newline$ $g(\pi) = \frac{0}{2}$ $\newline$ $g(\pi) = 0$
Calculate Limit: Since we get a defined value $0$ when we substitute $x = \pi$ , the limit exists and we have found it.