Is the function bounded each petal consider to be continuous when n tend to infinity? Was plotting this graph in matlab on the cartesian axis(x,y) and $0<t<2\pi$,this function will generate 2n "petal" depend on the input number n.
$x=r.cos(t)$
$y=r.sin(t)$
$r=2.sin(nt)$
As example n=10, noticed the curve bounded each petal is continuous and by using chain rule, the derivative found
$\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{n.sin(t).cos(nt)+sin(nt).cos(t)}{n.cos(t).cos(nt)-sin(nt)sin(t)}$

Then I consider n=1000 which is what I get

After enlarge the edge, noticed each curve as if squeeze into a pointy line

When n is large, the derivative become
$\frac{dy}{dx}=\frac{\frac{n.sin(t).cos(nt)+sin(nt).cos(t)}{n}}{\frac{n.cos(t).cos(nt)-sin(nt)sin(t)}{n}}\approx tan(t)$
Now I am confused regarding the concept of continuity, I read on internet for a function f(x) if it is differentiable at c then it is continuous at C but at the same time, from internet I also get to know "pointy line" is not differentiable. So for this function when n tend to infinity, the line bounded each petal still consider to be continuous? as its derivative exist but at the same time each of them is a pointy line?
 A: As Ivan Neretin has pointed out, you are only seeing "pointy lines" because you haven't enlarged enough. Magnify more and you will eventually see smooth turn-arounds instead of points, similar to what you see in the $n =10$ graph, though with straighter sides.
For any finite $n$ you will find the same holds. The higher $n$ is, the smaller the radius-of-curvature of the turn-around, and thus the more magnification will be needed to not see it as a point, but that radius-of-curvature will still be $> 0$, so at some magnification, it will appear smooth.
What happens when you let $n \to \infty$? You cease to have a curve. $r = \sin n\theta$. If $n = \infty$, this no longer makes sense. For a fixed angle $\theta$, as $n \to \infty$ the values of $r$ do not approach a single limit, but many different limit points. When $\theta$ is an irrational multiple of $\pi$, $r$ will approach every value in $[-1,1]$. For rational multiples of $\pi$, there will be only a finite number of limit points. Letting $n \to \infty$ does not result in a curve, so no results about the behavior of curves are applicable.
