showing that a function has a weak tangent but no strong tangent at t = 0

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

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

so for the case of a weak tangent, i have that and . so

in the case for a strong tangent i have and . then

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?

Re: showing that a function has a weak tangent but no strong tangent at t = 0

Quote:

Originally Posted by

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

let

be a continuous and simple (one to one) curve. we say

has a weak tangent at

if the line determined by

and

has a limit position as h approaches 0. we say

has a strong tangent at

if the line determined by

and

has a limit position as h, k approach 0. show that

has a weak tangent but no strong tangent at t = 0.

so for the case of a weak tangent, i have that

and

. so

in the case for a strong tangent i have

and

. then

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?

The limits that you have calculated are not the relevant ones. The weak and strong tangents are the limiting positions of *lines*, not *points*. What you should be looking at is whether the gradient of the line joining the given points tends to a limit.

The gradient of the line joining to is , which tends to infinity as h goes to 0. So the weak tangent is the vertical line through the origin.

The gradient of the line joining to is If for example k=–h then that fraction is 0. Use that fact to convince yourself that the limit as does not exist.