I want to compute $$\lim\limits_{n \to \infty}\int_{0}^{1}(\cos x-\sin x)^n dx.$$
I tried the squeeze theorem, tried simplifying the integral, but I eventually got nothing.
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And the differential? – Tito Eliatron May 29 '19 at 18:13
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@Clayton The bounds are definitely $0$ and $1$ – Rareș Stanca May 29 '19 at 18:21
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HAve you tried Lebesgue Dominated Convergene Theorem? – Tito Eliatron May 29 '19 at 18:25
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it looks like it's bounded convergence – user29418 May 29 '19 at 18:27
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@TitoEliatron Unfortunately, I haven't learned that yet. – Rareș Stanca May 29 '19 at 18:27
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WHat is the background of your question? What do you know? Uniform convergence of sequences? Only elemental calculus? – Tito Eliatron May 29 '19 at 18:29
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@TitoEliatron Only elemental calculus. – Rareș Stanca May 29 '19 at 18:29
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the graph grows bounded in the interval so it should be 0. – user29418 May 29 '19 at 18:41
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Because $(\cos(x)-\sin(x)) \leq 1-x$ over the range of integration, an upper bound for the integral is $$\int_{0}^{1} (1-x)^n dx.$$ But, $$\lim_{n \rightarrow \infty} \int_{0}^{1} (1-x)^n dx = \lim_{n \rightarrow \infty} \frac{1}{n+1} = 0.$$
Because the original integral is positive for all $n$, the limit must be zero by the Squeeze theorem.
D.B.
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i guessed the correct method! I thought it would be bounded by (1-x)^n – user29418 May 29 '19 at 18:42
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1I would assume that the the statement "the original integral is positive" needs a proof. – user May 29 '19 at 19:57
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Observe $$\int_0^1f(x)^n\,dx=\int_0^\varepsilon f(x)^n\,dx+\int_\varepsilon^1 f(x)^n\,dx$$where $f(x)=\cos(x)-\sin(x)$. Now, for $0\leq x\leq 1$, we have $|\cos(x)-\sin(x)|\leq1$ with equality only when $x=0$. Then the first integral can be trivially estimated above by $\varepsilon$ and the second integral can be estimated above by $\cos(\varepsilon)^n(1-\varepsilon)<\cos^n(\varepsilon)$. Since $\cos(\varepsilon)<1$, this limit must go to $0$. In particular, for an arbitrary $\delta>0$, we can take $n$ large enough and $\varepsilon$ small enough so that $$\left|\int_0^1 f(x)^n\,dx\right|<\delta.$$This formally proves that the limit must equal $0$.
Clayton
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