Formalizing using $\sin h \approx h$ to evaluate a limit.

Question

Assume $f \in C^{\infty}(\mathbb{R}^2)$ satisfies $f(\sin 2t, \sin t)=0$ for all $t\in \mathbb{R}$. I want to evaluate the limit \begin{equation*} \lim_{h \to 0}\frac{f(2h,h)}{h}. \end{equation*} It seems that (from the context from which this question came up) this limit should be equal to $0$ and an intuitive rationale that supports this is to consider $h\approx \sin h$, so that if we rewrite the limit above as \begin{equation*} \lim_{\theta \to 0}\frac{f(2\sin\theta,\sin \theta)}{\sin\theta} \underset{\text{looks like}}{=}\hspace{0.1cm} \lim_{\theta \to 0}\frac{f(2\sin\theta\cos \theta,\sin \theta)}{\sin\theta} = 0, \end{equation*} where the last equality follows from the fact that $f$ is $0$ in points of the form $(\sin2t, \sin t)$. How to formalize this intuition?

Answer 1

$f(u,v)=au+bv+o(\|(u,v)\|)$ gives

$f(\sin(2t),\sin t)=a(2t+o(t))+b(t+o(t))+o(t)=(2a+b)t+o(t)$.

Since $f(\sin(2t),\sin t)=0$, we deduce $2a+b=0$ hence

$f(2h,h)=a(2h)+bh+o(h)=o(h)$.

Answer 2

We can evaluate

$\frac{d}{dt}f(\sin (2t), \sin(t)) = (\frac{\partial f}{\partial x} 2 \cos(2t) + \frac{\partial f}{\partial y} \cos(t))|_{(x, y) = (2t, t)} = 0$

using the multivariable chain rule. Plugging in $t = 0$ gives us $(2 \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y})_{(x, y) = (0, 0)} = 0$.

Now note that $\lim\limits_{h \to 0} \frac{f(2h, h)}{h} = \frac{d}{dt} f(2t, t)|_{t = (0, 0)}$ by the definition of the derivative. We apply the multivariable chain rule again to get that this derivative is $(2 \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y})_{(x, y) =(0, 0)}$. But we’ve already demonstrated that this quantity is $0$.

Answer 3

You would want to be careful with intuition such as this. For example, suppose you defined a function $f$ by $$f(x,y) = \frac{x - 2y\sqrt{1-y^2}} {y^3}.$$ Then $f(\sin(2\theta), \sin(\theta)) = 0$ at least for $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$. However, $f(2h, h) = \frac{2h - 2h\sqrt{1-h^2}}{h^3}$, and a Taylor expansion of the square root term will show that $f(2h, h) \to 1$ as $h\to 0$, so $\frac{f(2h, h)}{h}$ would have no limit.

Of course, this has the issue that $f(x,y)$ is not defined on the $x$-axis. You could try making some adjustments near $y=0$ to get a better defined function. However, if you do that, you might see that you continue to get functions which are not differentiable at the origin -- as the following argument using the differentiability will show:

Under the given conditions, in particular, $f$ is continuous at $(0, 0)$. Therefore, $(\sin(2t), \sin(t)) \to (0, 0)$ as $t\to 0$ implies $f(0, 0) = \lim_{t\to 0} f(\sin(2t), \sin(t)) = 0$. Also, using the multivariate chain rule, we have that $$\left. \frac{d}{dt} [f(\sin(2t), \sin(t))] \right|_{t=0} = 2 f_x(0, 0) + f_y(0, 0) = 0.$$ Now since $f(0, 0) = 0$, the given limit is equal by definition to $$\left. \frac{d}{dt} [f(2t, t)] \right|_{t=0} = 2 f_x(0, 0) + f_y(0, 0) = 0.$$

Answer 4

For $h\in[-1,1]$ you get $f(2h\sqrt{1-h^2},h)=0.$

So $$\frac{f(2h,h)}{h}=\frac{f(2h,h)-f(2h\sqrt{1-h^2},h)}h\tag1$$

Now, by the mean value theorem, there is some $h'$ between $h$ and $h\sqrt{1-h^2}$ so that $$f(2h,h)-f(2h\sqrt{1-h^2},h)=2h\left(1-\sqrt{1-h^2}\right) f_x(2h',h).$$

So, if $f_x$ is bounded by $C$ in some neighborhood of $(0,0),$ then:

$$\left|\frac{f(2h,h)-f(2h\sqrt{1-h^2},h)}h\right|\leq 2C\left(1-\sqrt{1-h^2}\right)$$

So the right hand side of $(1)$ goes to zero as $h\to 0.$

So you only need that $\frac{\partial f}{\partial x}$ exists and is bounded on some neighborhood of $(0,0).$