I have some confusion regarding the existence of the follow limit:
$$\lim _{a \rightarrow \infty} \frac{\int_0^a \sin ^4 x d x}{a}$$
According to the answer key of the material I found this question in (from a local study institute), this limit does not exist.
I also verified this using Desmos, and it is true that as $a$ gets large the function does not appear to take a constant value.
However this is how I solved it:
Let $$a=n\frac{\pi}{2}+k , 0<k<\frac{\pi}{2},n\in \mathbb{N}$$
Then we can rewrite the limit as $$\lim_{n \rightarrow \infty}\frac{\int_{0}^{n\frac{\pi}{2}+k}\sin^4x\ dx}{n\frac{\pi}{2}+k}$$ $$\lim_{n \rightarrow \infty}\frac{\int_{0}^{n\frac{\pi}{2}}\sin^4x\ dx + \int_\frac{n\pi}{2}^{n\frac{\pi}{2}+k}\sin^4x\ dx}{n\frac{\pi}{2}+k}$$ The $k$ in the denonimator can be neglected with respect to $n\frac{\pi}{2}$, and the first integral evaluates to $\frac{3n\pi}{16}$, so we get
$$\lim_{n \rightarrow \infty}\frac{\frac{3n\pi}{16}+\int_\frac{n\pi}{2}^{n\frac{\pi}{2}+k}\sin^4x\ dx }{\frac{n\pi}{2}}$$
The second integral can be neglected in comparison the first so finally we get $$\lim_{n \rightarrow \infty}\frac{\frac{3n\pi}{16}}{\frac{n\pi}{2}}=\frac{3}{8}$$
And yet, this limit $\it{apparently}$ does not exist. So, can someone tell me what's wrong with my method and $\it{why}$ does the limit not exist?
Best Answer
The limit does exist. A slightly more general statement is as follows:
In OP's case, this theorem implies that
$$ \lim_{a \to \infty} \frac{1}{a} \int_{0}^{a} \sin^4 x \, \mathrm{d}x = \frac{1}{\pi} \int_{0}^{\pi} \sin^4 x \, \mathrm{d}x = \frac{3}{8} $$
as is numerically verified from the below plot of $a \mapsto \frac{1}{a} \int_{0}^{a} \sin^4 x \, \mathrm{d}x$:
This can also be directly verified by using the formula
$$ \int_{0}^{a} \sin^4 x \, \mathrm{d}x = \frac{3a}{8} - \frac{\sin(2a)}{4} + \frac{\sin(4a)}{32}. $$
Proof of Theorem. Let $\overline{f} = \frac{1}{T} \int_{0}^{T} f(x) \, \mathrm{d}x$ denote the "average value" of $f$, and define
$$ F(a) = \int_{0}^{a} f(x) \, \mathrm{d}x - a \overline{f} . $$
The assumption tells that $F(\cdot)$ is continuous and $T$-periodic. In particular, $F$ is bounded on all of $\mathbb{R}$. Therefore we have
$$ \lim_{a\to\infty} \frac{F(a)}{a} = 0, $$
which is equivalent to the assertion of the theorem. $\square$