Tangent line limit of a slope

Question

I want to find the equation of the tangent line to $x(t)=2\cos(t)\cos(2t)$, $y(t)=2\sin(t)-\sin(2t)$ at $t=0$.

I used this site https://www.emathhelp.net/calculators/calculus-1/tangent-line-calculator/?ft=t&f=2cos%28t%29cos%282t%29&g=%272sin%28t%29-sin%282t%29&px=0

According to it, the equation of the tangent line is $y=0$. Looking at the solution, I noticed that to find the slope, the limit is taken. Is it correct to do so, or should I just plug the values and say that the slope is undefined, so the tangent line doesn't exist?

A limit tells how the function behaves near a point, not at it. The tangent line should be found at the point, not near it.

Is there a reason for the taking of the limit to be justified?

Thank you.

Answer 1

I agree with OP that the tangent line should be found at the point in question, and a justification is required for finding the slope at a nearby point and then taking the limit of that slope as that nearby point approaches the point in question.

Let $P$ be the point $(x(0), y(0)) = (2,0)$. If you graph the curve, you will find that it has a cusp at $P$, so it is not clear to me that the there is a line that should be called the tangent line at $P$. Nevertheless, the following calculation gives the line that is tangent to both sides of the cusp at $P$.

The usual way to find the slope of the tangent line at $P$ is to find the slope of the line through $P$ and a nearby point $Q$ on the curve, and then take the limit of this slope as $Q$ approaches $P$. The nearby point $Q$ will be $(x(t), y(t))$ for $t \ne 0$, and the slope of the line through $P$ and $Q$ is $$ \frac{y(t) - 0}{x(t) - 2} = \frac{2\sin(t) - \sin(2t)}{2\cos(t)\cos(2t) - 2}. $$ So the tangent slope we want is $$ \lim_{t \to 0} \frac{2\sin(t) - \sin(2t)}{2\cos(t)\cos(2t) - 2}. $$ By L'Hopital's rule, we can find this limit by computing $$ \lim_{t \to 0} \frac{\cos(t) - \cos(2t)}{6\sin^3(t)-5\sin(t)}. $$ This is the same as the limit on the site referred to by OP, but now a justification has been given for computing that limit.

Answer 2

Look at the step where $M(t_0)$ has been derived. One can observe that $M(t_0)$ is a rational function whose denominator is $0$ when $t = 0$. So, $M(0)$ does not exist. This is what justifies taking the limit as $t \to 0$. Taking the limit allows you to apply L'Hopital's rule, which requires some more justification than I've given here.

Answer 3

At the point under consideration the slope is continuous, passing from negative to positive via zero.

So at this point the tangent is the x-axis, $y=0$. ( Like a corner of the astroid on x-axis).