f(x)=(2-cos^(2)(x))/(sin(2x)) We want to find lim_(x rarr(pi)/(2))f(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 = π/2 into the function f(x) = (2 - cos^2(x)) / sin(2x) to see what we get. f(π/2) = (2 - cos^2(π/2)) / sin(2 * π/2)Calculate Values: Now we calculate the cosine and sine values at x = π/2. cos(π/2) = 0, so cos^2(π/2) = 0^2 = 0. sin(π/2) = 1, so sin(2 * π/2) = sin(π) = 0.Substitute Values: Substitute these values into the function. f(π/2) = (2 - 0) / 0 f(π/2) = 2 / 0 This results in a division by zero, which is undefined.Division by Zero: Since we have a division by zero, the limit does not exist through direct substitution, and we encounter an indeterminate form of the type 2/0, which suggests a possible asymptote or that the limit does not exist.

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f(x)=(2-cos^(2)(x))/(sin(2x))
We want to find
lim_(x rarr(pi)/(2))f(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.

$f(x)=\frac{2-\cos ^{2}(x)}{\sin (2 x)}$ $\newline$ We want to find $\lim _{x \rightarrow \frac{\pi}{2}} f(x)$ . $\newline$ 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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Algebra 1

Domain and range of relations

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f(x)=\frac{2-\cos ^{2}(x)}{\sin (2 x)}

\newline

We want to find

\lim _{x \rightarrow \frac{\pi}{2}} f(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 $x = \frac{\pi}{2}$ into the function $f(x) = \frac{(2 - \cos^2(x))}{\sin(2x)}$ to see what we get. $f\left(\frac{\pi}{2}\right) = \frac{(2 - \cos^2\left(\frac{\pi}{2}\right))}{\sin(2 \cdot \frac{\pi}{2})}$
Calculate Values: Now we calculate the cosine and sine values at $x = \frac{\pi}{2}$ . $\cos(\frac{\pi}{2}) = 0$ , so $\cos^2(\frac{\pi}{2}) = 0^2 = 0$ . $\sin(\frac{\pi}{2}) = 1$ , so $\sin(2 \cdot \frac{\pi}{2}) = \sin(\pi) = 0$ .
Substitute Values: Substitute these values into the function. $\newline$ $f(\frac{\pi}{2}) = (2 - 0) / 0$ $\newline$ $f(\frac{\pi}{2}) = \frac{2}{0}$ $\newline$ This results in a division by zero, which is undefined.
Division by Zero: Since we have a division by zero, the limit does not exist through direct substitution, and we encounter an indeterminate form of the type $\frac{2}{0}$ , which suggests a possible asymptote or that the limit does not exist.