I was presented the following problem at Calculus class:
Let $f: (0, \infty) \to \mathbb{R}$. We know that $f$ is Riemann integrable in every compact interval of $(0, \infty)$, and that the limit $\underset{t \to 0^+}{\lim} f(t) = L$ exists. Show that for all $0 < \alpha < \beta$ we have $$\lim_{t \to 0^+}\int \limits_{\alpha t}^{\beta t} \frac{f(x)}{x}dx = L \log\frac{\beta}{\alpha}.$$ I showed that for every $\varepsilon > 0$ there exists $T>0$ such that $$(L-\varepsilon)\log\frac{\beta}{\alpha}< \int \limits_{\alpha t}^{\beta t} \dfrac{f(x)}{x}dx < (L+\varepsilon)\log\frac{\beta}{\alpha}.$$ Indeed, it is easy to show using the given limit of $f$: there exists $T>0$ such that for every $0 < t < T$ we have $L-\varepsilon < f(x) < L+\varepsilon$. Then, for every $\beta t < T$ $$\int \limits_{\alpha t}^{\beta t} \frac{f(x)}{x}dx<\int \limits_{\alpha t}^{\beta t} \frac{L + \varepsilon}{x}dx =\cdots = (L+\varepsilon)\log\frac{\beta}{\alpha}$$
and symmetrically for the lower bound. My professor said that this proof does not show what is needed to be shown, however I cannot see why this is not correct. As I see it, this is exactly what is needed to be shown, following the definition of the limit, i.e.
$$\lim_{t \to 0^+} \int \limits_{\alpha t}^{\beta t} \dfrac{f(x)}{x}dx = L \log\frac{\beta}{\alpha}\iff(L-\varepsilon)\log\frac{\beta}{\alpha}< \int \limits_{\alpha t}^{\beta t} \frac{f(x)}{x}dx < (L+\varepsilon)\log\frac{\beta}{\alpha}.$$
Am I missing something? I will appreciate it if someone could kindly shed a light on this matter.
Since you hand us the stick to beat you, I shall be (over?)demanding. Ready to hear (unkind?) criticism?
You write "for every $\varepsilon > 0$ there exists $T>0$ such that", followed by some things where $T$ never occurs. The missing piece inbetween is "$\forall t\in(0,T)$". Edit: since you edited your proof, this problem partially disappeared but there is still a $\forall t$ missing just after "there exists $T>0$ such that".
What I find more embarrassing is that you stop at $(L-\varepsilon)\log\frac\beta\alpha<\int_{\alpha t}^{\beta t}\frac{f(x)}xdx<(L+\varepsilon)\log\frac\beta\alpha$ and pretend the proof is done. It is not, because your definition of the limit suffers the same sloopiness and lack of quantifiers. Indeed, $\lim_{t\to0^+}\int_{\alpha t}^{\beta t}\frac{f(x)}xdx=L\log\frac\beta\alpha$ is not equivalent to "$(L-\varepsilon)\log\frac\beta\alpha<\int_{\alpha t}^{\beta t}\frac{f(x)}xdx< (L+\varepsilon)\log\frac\beta\alpha$" but to $$\forall\epsilon>0\quad\exists\delta>0\quad\forall t\in(0,\delta)\quad L\log\frac\beta\alpha-\epsilon<\int_{\alpha t}^{\beta t}\frac{f(x)}xdx<L\log\frac\beta\alpha+\epsilon.$$ So, what is missing here is the choice of $\delta,$ given $\epsilon$ (of course, it will rely on your previous $\forall\varepsilon\exists T\dots$, but some work remains to be done).
Admittedly, this missing piece is boaring to write and you seem to have the right feeling, but your professor probably does not find it sufficient.