The following problem is from Chapter 14 "The Fundamental Theorem of Calculus" from Spivak's Calculus
- (a) If $f$ is continuous on $[0,1]$, compute $\lim\limits_{x\to 0^+} x\int_x^1 \frac{f(t)}{t}dt$.
I am aware that someone asked a question about this problem but the question I have is not about the use of L'Hôpital's Rule or about a solution involving supremums or logarithms. I am interested in a proof along the lines of the solution manual, which uses knowledge present in the book only up to this chapter.
Here is my solution which provides many intermediate steps that the solution manual doesn't. I'd like to know if they are correct.
We can rewrite the limit as
$$\lim\limits_{x\to 0^+} \frac{\int_x^1 \frac{f(t)}{t}dt}{\frac{1}{x}}$$
Let
$$f(x)=\int_x^1 t^{-1}dt$$
$$g(x)=x^{-1}$$
It can be shown that
$$\lim\limits_{x\to 0^+} f(x)=\lim\limits_{x\to 0^+} g(x)=\infty$$
$$\lim\limits_{x\to 0^+} x^{-1}=\infty$$
Also,
$$\lim\limits_{x\to 0^+} \frac{f'(x)}{g'(x)}= \lim\limits_{x\to 0^+} \frac{-\frac{1}{x}}{-\frac{1}{x^2}}=\lim\limits_{x\to 0^+} x=0$$
Therefore, by L'Hôpital's Rule we can infer that
$$\lim\limits_{x\to 0^+} x \int_x^1 t^{-1} dt=\lim\limits_{x\to 0^+} \frac{f(x)}{g(x)}=\lim\limits_{x\to 0^+} \frac{f'(x)}{g'(x)}=0\tag{1}$$
But we want to compute $\lim\limits_{x\to 0^+} x\int_x^1 \frac{f(t)}{t}dt$.
Since $f$ is continuous on $[0,1]$, it is bounded on that interval. Therefore, there is some $M>0$ such that for $x\in [0,1]$ we have $|f(x)|\leq M$.
Therefore
$$-M \leq f(x)\leq M$$
$$-\frac{M}{x}\leq \frac{f(x)}{x}\leq \frac{M}{x}$$
$$-\int_x^1 \frac{M}{t}dt\leq \int_x^1\frac{f(t)}{t}dt\leq \int_x^1\frac{M}{t}dt$$
Multiply by $x$ (and remember that $x\in [0,1]$
$$-x\int_x^1 \frac{M}{t}dt\leq x\int_x^1\frac{f(t)}{t}dt\leq x\int_x^1\frac{M}{t}dt$$
And take the limit
$$0=\lim\limits_{x\to 0^+}\left [-x\int_x^1 \frac{M}{t}dt\right ]\leq \lim\limits_{x\to 0^+}x\int_x^1\frac{f(t)}{t}dt\leq \lim\limits_{x\to 0^+} x\int_x^1\frac{M}{t}dt=0$$
Hence we have
$$\lim\limits_{x\to 0^+}x\int_x^1\frac{f(t)}{t}dt=0$$
$$\blacksquare$$
The solution manual has something very similar, but as usual is very terse and skips many intermediate steps. I'd like to know if my solution above is correct at every step.
Best Answer
You’re basically right, except that you’ve used the notation $f$ twice to denote two different functions. You can use $F(x)=\int_x^1 t^{-1}dt$ to avoid this ambiguity. Another minor remark: note that $\int_0^1 t^{-1}\,dt=\infty$, so all inequalities (except $|f(x)|\leq M$) are only valid for $x\in(0,1]$. If you want to write a very rigorous proof, you need to watch out on the validity of your every inequality.
A quicker way to prove $\lim\limits_{x\to 0^+} x \int_x^1 t^{-1} dt=0$: for all $x\in(0,1)$ we have $$\int_x^1 t^{-1} dt=\log t|_x^1=-\log x,$$ hence $$\lim\limits_{x\to 0^+} x \int_x^1 t^{-1} dt= -\lim\limits_{x\to 0^+} x\log x=0.$$