I'm trying to solve some Do Carmo problems from his book Differential geometry of curves and surfaces. In section 1-3 prob.5.c., we have the curve: $\alpha:(-1,\infty)\rightarrow\Bbb R^2$ given by: $$\alpha(t)=(\frac{3at}{1+t^3},\frac{3at^2}{1+t^3})$$ And they ask: take the curve with the opposite orientation. Now, as $t\to-1$, the curve and its tagent approach the line $x+y+a=0$.
I wasn't really sure of what was supose to hapen, but I took the derivative of $\alpha$: $$\alpha'(-t)=(\frac{3a(1-t^2+3t^3)}{(1+t^3)^2},\frac{-3at(2-2t^3-3t^4)}{(1+t^3)^2})$$ And then I looked for the limit if $\alpha,\alpha'$ when $t\to -1$ to see what happened, but I got that they both tend to constants as spected since both are continuous, so what happened? What am I supose to see?
