Let's consider the following question
Let $x=\frac{t}{1+t^2},$ $y=3t^2-6t$. Find $\frac{dy}{dx}\Big|_{t=1}$.
It is known that $$\frac{dy}{dx}=\frac{dx/dt}{dy/dt}.$$
The problem is that both $\frac{dy}{dt}\Big|_{t=1}=0$, and $\frac{dy}{dt}\Big|_{t=1}=0$. In this case, we can show that
\begin{equation}\label{a} \frac{dy}{dx}\bigg|_{t=1}=\lim_{t\to1}\frac{dy}{dx}. \end{equation} My question: If $\lim_{t\to1}\frac{dy}{dx}$ exists, is it necessarily true that $\frac{dy}{dx}\Big|_{t=1}=\lim_{t\to1}\frac{dy}{dx}$. ?
Thanks
Let $f$ be a continuous function at $a$, and let it be differentiable on some punctured neighbourhood of $a$, with $\lim_{x\to a}f'(x)$ defined.
Given L'Hôpital's rule, we have that:
$f'(a):=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{x\to a} f'(x)$
In case we are talking about a curve and not of a function, the same way of thinking can be applied to components to obtain the desired result