Is the identity true?

Question

Is the following identity true:

$$\lim\limits_{\Delta x \to 0} \dfrac{\displaystyle \int_0^{\Delta x} f(x)\ dx}{\displaystyle \int_0^{\Delta x} g(x)\ dx} =\dfrac{\lim\limits_{\Delta x \to 0} \displaystyle \int_0^{\Delta x} f(x)\ dx}{\lim\limits_{\Delta x \to 0} \displaystyle \int_0^{\Delta x} g(x)\ dx} =\dfrac{f(x) \lim\limits_{\Delta x \to 0} \displaystyle\int_0^{\Delta x} dx} {g(x) \lim\limits_{\Delta x \to 0} \displaystyle\int_0^{\Delta x} dx} =\dfrac{f(x)}{g(x)}$$

If yes:

$(1)$ How is taking $f(x)$ and $g(x)$ out of the integral mathematically justifyable?

$(2)$ Can we extend this integral to higher integrals (surface and volume integrals)? For example:

$$\lim\limits_{\Delta V \to 0} \dfrac{\displaystyle \int_{\Delta V} f(x,y,z)\ dV}{\displaystyle \int_{\Delta V} g(x,y,z)\ dV}$$

Answer 1

Well yes, I think what you meant is correct, but not what you wrote.

The first equality is only true if both limits exist, are finite and the denominator is non-zero, which is not true here. Then the second equality doesn't make sense, as $x$ is a bound variable and is not defined outside the integral, hence leading to the inconsistency that @QtizedQ pointed out. Finally, it should be clear that $\lim_{\Delta x\to 0}\int_0^{\Delta x}dx=0$ and so, in your last equality, you're dividing both the numerator and denominator by $0$.

However, what is true (assuming that $f$ and $g$ are integrable with $g$ non-zero) is that

$$\lim_{\Delta x\to0}\frac{\int_{x_1}^{x_1+\Delta x}f(x')dx'}{\int_{x_2}^{x_2+\Delta x}g(x')dx'}=\frac{f(x_1)}{g(x_2)}$$

You can show this pretty easily, using the Fundamental Theorem of Calculus. For that, let $$F(x):=\int_0^xf(x)dx\quad\text{and}\quad G(x):=\int_0^xg(x)dx$$

Then $$\lim_{\Delta x\to0}\frac{\int_{x_1}^{x_1+\Delta x}f(x')dx'}{\int_{x_2}^{x_2+\Delta x}g(x')dx'}=\lim_{\Delta x\to0}\frac{\frac{\int_{x_1}^{x_1+\Delta x}f(x')dx'}{\Delta x}}{\frac{\int_{x_2}^{x_2+\Delta x}g(x')dx'}{\Delta x}}=\lim_{\Delta x\to0}\frac{\frac{F(x_1+\Delta x)-F(x_1)}{\Delta x}}{\frac{G(x_2+\Delta x)-G(x_2)}{\Delta x}}=\frac{F'(x_1)}{G'(x_2)}=\frac{f(x_1)}{g(x_2)}$$

where the last equality comes from the Fundamental Theorem of Calculus.

As for the multivariable case, you can just apply the above result successively. If $\Delta V$ is a neighborhood of $(x,y,z)$, you can take the limit where its volume goes to $0$: \begin{align} \lim_{\text{vol}\left(\Delta V\right)\to0}\frac{\int_{\Delta V}f(x',y',z')dV}{\int_{\Delta V}g(x',y',z')dV}&=\lim_{\Delta x\to0, \Delta y\to0, \Delta z\to0}\frac{\int_x^{x+\Delta x}\int_y^{y+\Delta y}\int_z^{z+\Delta z}f(x',y',z')dx'dy'dz'}{\int_x^{x+\Delta x}\int_y^{y+\Delta y}\int_z^{z+\Delta z}g(x',y',z')dx'dy'dz'}\\ &=\lim_{\Delta x\to0, \Delta y\to0}\frac{\int_x^{x+\Delta x}\int_y^{y+\Delta y}f(x',y',z)dx'dy'}{\int_x^{x+\Delta x}\int_y^{y+\Delta y}g(x',y',z)dx'dy'}\\ &=\lim_{\Delta x\to0}\frac{\int_x^{x+\Delta x}f(x',y,z)dx'}{\int_x^{x+\Delta x}g(x',y,z)dx'}=\\ &=\frac{f(x,y,z)}{g(x,y,z)} \end{align}

Answer 2

No. Let $f(x):=\tan(x)$ and $g(x):=\cos(x)$. Then

$$\lim\limits_{\Delta x \to 0} \dfrac{\displaystyle \int_0^{\Delta x} f(x)\ dx}{\displaystyle \int_0^{\Delta x} g(x)\ dx} =\lim\limits_{\Delta x \to 0} \dfrac{f(\Delta x)}{g(\Delta x)}=0$$

But $f(x)/g(x)\not\equiv 0$. Taking $f$ and $g$ out of the integral is not mathematically justified.

Answer 3

In the equations you've written:

• The first equality is unjustified as the limits in the numerator and denominator are both zero. So this would be one of those “indeterminate forms” for a limit.

• The second equality is unjustified as well. In the integral, $x$ is a dummy variable so $\int_0^{\Delta x} f(x)\,dx$ is a real number for each $\Delta x$. But $f(x)\int_0^{\Delta x}\,dx$ is a function of $x$; in fact, it's equal to $f(x)\,\Delta x$.

But you can use L'Hôpital's rule and the Fundamental Theorem of Calculus here. Assume that $f$ is continuous on an open interval containing $0$, and $g(0) \neq 0$. Let $F(z) = \int_0^z f(x)\,dx$ and $G(z) = \int_0^z g(x)\,dx$. Then: \begin{align*} \lim_{\Delta x\to 0} F(\Delta x) &= 0 \\ \lim_{\Delta x\to 0} G(\Delta x) &= 0 \\ \lim_{\Delta x\to 0} F'(\Delta x) &= \lim_{\Delta x \to 0} f(\Delta x) = f(0) \\ \lim_{\Delta x\to 0} G'(\Delta x) &= \lim_{\Delta x \to 0} g(\Delta x) = g(0) \\ \end{align*} Therefore, the quotient $\frac{F'(\Delta x)}{G'(\Delta x)}$ tends to $\frac{f(0)}{g(0)}$ as $\Delta x\to 0$. So by L'Hôpital's rule, $\frac{F(\Delta x)}{G(\Delta x)}$ has the same limit. In summary, $$ \lim_{\Delta x \to 0} \frac{\int_0^{\Delta x}f(x)\,dx}{\int_0^{\Delta x} g(x)\,dx} = \frac{f(0)}{g(0)}. $$

The multivariable analogue is a bit trickier. By the integral mean value theorem, the integral of a (continuous) function over a region is equal to the volume of that region times the value of a function at some point in the region. But with two integrals over the same region, it's not always possible to use the same point. So the quotient may not have a well-defined limit.

Answer 4

Both are false in the manner you have stated them, for example see the comment about the RHS being a function of $x$, while the LHS is a constant. FOr your higher diemnsional geenralisation, when you write $\lim_{\Delta V \to 0},$ you're thinking of $\Delta V$ as a number, but when you write $\int_{\Delta V} (\cdots)$, you're thinking of $\Delta V$ as a subset of $\Bbb{R}^3$. However, your intuition seems to be correct.

I'll offer you two possible statements which can be proven easily. Note that I make no attempt at minimal hypothesis... so you could probably get away with weaker hypotheses, but then necessarily the proof will have to be more involved.

• Suppose $f,g: \Bbb{R} \to \Bbb{R}$ are continuous functions, and that there is an interval containing $0$, where $g$ doesn't vanish. ($\exists \delta > 0$ such that $ \forall x\in (- \delta, \delta)$, $g(x) \neq 0$). Then, it is true that \begin{equation} \lim_{h \to 0} \dfrac{\displaystyle\int_0^hf}{\displaystyle\int_0^hg} = \dfrac{f(0)}{g(0)}. \end{equation}

In higher dimensions:

• Let $f,g: \Bbb{R}^n \to \Bbb{R}$ be continuous functions, and suppose that there is an open ball around the origin $0$, where $g$ doesn't vanish. For $r > 0$, let $B_r$ denote the closed ball of radius $r$ around the origin. Then, \begin{equation} \lim_{r \to 0} \dfrac{\displaystyle\int_{B_r}f}{\displaystyle\int_{B_r}g} = \dfrac{f(0)}{g(0)} \end{equation}

  ---

In the single variable case, you can see this as an application of LHopital's rule and the fundamental theorem of calculus. But I think using the Mean-Value theorem for integrals is more easily generalised. So, I'll prove it using that instead.

Note that for any $h \in \Bbb{R}$, (since $f$ and $g$ are continuous) we can apply the mean value theorem of integrals to say \begin{equation} \int_0^h f = f(\xi_h) \cdot h \quad \text{and} \quad \int_0^h g = g(\eta_h) \cdot h \end{equation} for some $\xi_h, \eta_h$ between $0$ and $h$. So, \begin{align} \lim_{h \to 0} \dfrac{\displaystyle\int_0^hf}{\displaystyle\int_0^hg} &= \lim_{h \to 0} \dfrac{f(\xi_h) \cdot h}{g(\eta_h) \cdot h} \\ &= \lim_{h \to 0} \dfrac{f(\xi_h)}{g(\eta_h)} \\ &= \dfrac{f(\lim_{h \to 0} \xi_h)}{g(\lim_{h \to 0} \eta_h)} \tag{$*$}\\ &= \dfrac{f(0)}{g(0)}. \end{align} By the way in $(*)$, we can bring the limit inside the functions because we assumed continuity of $f$ and $g$.

In the multidimensional case, the exact same proof works, with some minor notational changes: for any $r> 0$, by the mean-value theorem for integrals, we can say that \begin{equation} \int_{B_r} f = f(\xi_r) \cdot \text{vol}(B_r) \quad \text{and} \quad \int_{B_r} g = g(\eta_r) \cdot \text{vol}(B_r) \end{equation} for some $\xi_r, \eta_r \in B_r$. Hence, \begin{align} \lim_{r \to 0} \dfrac{\displaystyle \int_{B_r}f}{\displaystyle \int_{B_r}g} &= \lim_{r \to 0} \dfrac{f(\xi_r) \cdot \text{vol}(B_r)}{g(\eta_r) \cdot \text{vol}(B_r)} \\ &= \lim_{r \to 0} \dfrac{f(\xi_r)}{g(\eta_r)} \\ &= \dfrac{f(\lim_{r \to 0} \xi_r)}{g(\lim_{r \to 0} \eta_r)} \tag{$**$} \\ &= \dfrac{f(0)}{g(0)} \end{align} Once again, we can bring the limits inside the functions in $(**)$ because we assumed continuity of the function $f$ and $g$.

(If you would like to see a proof of the mean-value theorem for integrals, let me know, I can add that into my post as well.)