Originally Posted by

**oblixps** in do carmo's differential geometry of curves and surfaces he has the following problem:

let $\displaystyle \alpha $ be a continuous and simple (one to one) curve. we say $\displaystyle \alpha $ has a weak tangent at $\displaystyle t_0 $ if the line determined by $\displaystyle \alpha (t_0 + h) $ and $\displaystyle \alpha (t_0) $ has a limit position as h approaches 0. we say $\displaystyle \alpha $ has a strong tangent at $\displaystyle t_0 $ if the line determined by $\displaystyle \alpha (t_0 + h) $ and $\displaystyle \alpha (t_0 + k) $ has a limit position as h, k approach 0. show that $\displaystyle \alpha(t) = (t^3, t^2) $ has a weak tangent but no strong tangent at t = 0.

so for the case of a weak tangent, i have that $\displaystyle \alpha(0) = (0, 0), \alpha(h) = (h^3, h^2) $ and $\displaystyle \alpha(h) - \alpha(0) = (h^3, h^2) $. so $\displaystyle \lim_{h \to 0} \frac{1}{h} (h^3, h^2) = \lim_{h \to 0} (h^2, h) = (0, 0) $

in the case for a strong tangent i have $\displaystyle \alpha(h) = (h^3, h^2), \alpha(k) = (k^3, k^2) $ and $\displaystyle \alpha(h) - \alpha(k) = (h^3 - k^3, h^2 - k^2) $. then $\displaystyle \lim_{h,k \to 0} \frac{1}{h-k} (h^3 - k^3, h^2 - k^2) = \lim_{h,k \to 0} (h^2 + hk + k^2, h + k) = (0, 0) $

but it seems to me that both have limit positions so both the weak and strong tangent exist at t = 0? i am not too clear on the definition he gives in the problem of weak and strong tangent. what does he exactly mean by "limiting position". if he means that the line will obtain a fixed "direction" then wouldn't neither a weak nor a strong tangent exist in this case since both lines in this case just go to (0, 0) so actually there isn't even a line at all, but just a point or the zero vector?