If fx-lim n- 1-sin 2 x n -n- N- then

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Continuity in an Interval

If fx=lim n...

Question

If $f (x) = {lim}_{n \to \infty} {(1 - {sin}^{2} x)}^{n}, n \in N$ , then

f (\frac{π}{4})

is nearly zero

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f (\frac{π}{4})

is equal to

\frac{1}{2}

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f (x)

is discontinuous at infinte number of points

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f (x)

is a periodic

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Solution

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Similar questions

f (x) = 1 + cos x + {cos}^{2} x + \dots \infty

f (x) + f (\frac{π}{2} + x) + f (\frac{π}{2} - x) + f (π - x)

f (x) = \int_{0}^{{sin}^{2} x} {sin}^{- 1} (\sqrt{t}) d t + \int_{0}^{{cos}^{2} x} {cos}^{- 1} (\sqrt{t}) d t

f (x) = 0

f (- π) = f (π) = 0

f (\frac{π}{2}) = - \frac{3 π^{2}}{4}

lim x \to - π \frac{f (x)}{sin (sin x)}

f (x) = \frac{1}{n} {[(n + 1) (n + 2) (n + 3) . . . (n + n)]}^{\frac{1}{n}}

{lim}_{n \to \infty} f (x)

\frac{4}{e}

lim n \to \infty \frac{1}{n} f (\frac{r}{n}) = \int_{0}^{1} f (x) d x

f (x)

x = 1

f^{2} (x) = 4 \forall x ϵ R

f (x)

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