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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)=2cos2(x)sin(2x) f(x)=\frac{2-\cos ^{2}(x)}{\sin (2 x)} \newlineWe want to find limxπ2f(x) \lim _{x \rightarrow \frac{\pi}{2}} f(x) .\newlineWhat happens when we use direct substitution?\newlineChoose 11 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.

Full solution

Q. f(x)=2cos2(x)sin(2x) f(x)=\frac{2-\cos ^{2}(x)}{\sin (2 x)} \newlineWe want to find limxπ2f(x) \lim _{x \rightarrow \frac{\pi}{2}} f(x) .\newlineWhat happens when we use direct substitution?\newlineChoose 11 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.
  1. Direct Substitution: Let's first try direct substitution of x=π2x = \frac{\pi}{2} into the function f(x)=2cos2(x)sin(2x)f(x) = \frac{2 - \cos^2(x)}{\sin(2x)}.f(π2)=2cos2(π2)sin(2π2)f\left(\frac{\pi}{2}\right) = \frac{2 - \cos^2\left(\frac{\pi}{2}\right)}{\sin(2 \cdot \frac{\pi}{2})}
  2. Substituting cos(π2)\cos(\frac{\pi}{2}): Now, we know that cos(π2)=0\cos(\frac{\pi}{2}) = 0, so cos2(π2)=02=0\cos^2(\frac{\pi}{2}) = 0^2 = 0. Substituting this into the function gives us: f(π2)=(20)/sin(π)f(\frac{\pi}{2}) = (2 - 0) / \sin(\pi)
  3. Division by Zero: Since sin(π)=0\sin(\pi) = 0, we have a division by zero situation.f(π2)=20f(\frac{\pi}{2}) = \frac{2}{0}This is undefined, which means we have an indeterminate form.
  4. Indeterminate Form: The indeterminate form we have is of the type 20\frac{2}{0}, which suggests that the limit does not exist because we cannot divide by zero.

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