Verification of my attempt to calculate $\lim_{x\to \infty} \frac{1}{x}\int_{2}^{x}(1-\frac{\cos^2{t}}{t})\, dt$ For calculating $$\lim_{x\to \infty} \frac{1}{x}\int_{2}^{x}\left(1-\frac{\cos^2{t}}{t}\right)\, dt$$
I have thought to apply the mean value theorem. In fact thanks to the continuity of $1-\frac{\cos^2{t}}{t}$ I can say that $\exists c\in[2, x]:$ $\int_{2}^{x}\left(1-\frac{\cos^2{t}}{t}\right)\, dt=f(c)(x-2)$.
So $$\lim_{x\to \infty} \frac{1}{x}\int_{2}^{x}\left(1-\frac{\cos^2{t}}{t}\right)\, dt=\lim_{x\to \infty} \frac{1}{x}f(c)(x-2)=1$$
since $f(c)=(1-\frac{\cos^2{c}}{c})=1$ because $c\to \infty$ when $x\to\infty$.
Is it right my work?
 A: It is not so evident that $c \to \infty$. However,  you could apply L'Hospital's rule to get
$$\lim_{x\to \infty} \frac{\int_{2}^{x}(1-\frac{\cos^2{t}}{t})\, dt}{x} = \lim_{x\to\infty} \frac{1-\frac{\cos^2{x}}{x}}{1} = 1$$
A: All we know for sure is that $c$ lies between $2$ and $x$ and hence we can't ensure that $c\to\infty $.
Mean value theorem can be used but it requires some more work in this context. Essentially note that integrand $f(t) \to 1$ as $t\to\infty$ and hence for any $\epsilon>0$ we have  number $M>0$ such that $1-\epsilon<f(t)<1+\epsilon$ if $t>M$.
Now split the interval into two $[2,M],[M,x]$ and then $$g(x) \frac{1}{x}\int_2^x f(t) \, dt=\frac {1}{x}\int_2^M f(t) \, dt+\frac{1}{x}\int_M^x f(t) \, dt$$ The first term on right tends to $0$ and the second term can be written as $$\frac{x-M} {x} f(c) $$ where $M<c<x$ so that $f(c) $ lies between $1-\epsilon $ and $1+\epsilon $. By taking limits as $x\to\infty$ we see that $$1-\epsilon \leq\liminf_{x\to \infty} g(x) \leq \limsup _{x\to\infty} g(x) \leq 1+\epsilon $$ Since $\epsilon>0$ was arbitrary we have $\lim_{x\to\infty} g(x) =1$.
A: Anytime $\lim_{t\to \infty} f(t) = L,$ with $f$ integrable on $[a,x]$ for all $x>a,$ we have
$$\tag 1 \lim_{x\to\infty}\frac{\int_a^x f(t)\,dt}{x} =L.$$
In our problem it is clear that $1-\dfrac{\cos^2 t}{t} \to 1.$ So the desired limit is $1.$
The proof of $(1)$ is very much like showing that if the sequence $a_k\to L,$ then $\dfrac{\sum_{k=1}^{n}a_k}{n}\to L.$
A: $f(x):=\displaystyle{\int_{2}^{x}} (1-(\cos^2 t)/t)dt$;
For $t \in [2,x] :$
$(1-1/t) \le (1-(\cos^2 t))/t \le 1;$
$(1/x)\displaystyle{\int_{2}^{x}}(1-1/t)dt \le (1/x)f(x) \le$
$(1/x)\displaystyle{\int_{2}^{x}}1dt;$
Take the limit.
Recall: $\lim_{x \rightarrow \infty} (1/x)\log x=0.$
