# Limit along the line

I came across the following passage:

Limit of a function $$f$$ along the line $$l: (x_0 +t\cos a, y_0+t\sin a)$$, $$t\in \mathbb{R}, a\in [0,\pi]$$, through the point $$(x_0, y_0)$$ is the following limit: $$\lim_{t\to 0}f(x_0 +t\cos a, y_0+t\sin a)$$

I don't understand the representation of the line $$l$$. Is the collection of points $$(x_0 +t\cos a, y_0+t\sin a)$$ a set of points that satisfies the equation of the line $$l$$ ? I'm familiar with polar coordinates and I've tried representing a point $$(x,y)$$ and plugging in the equation $$y=mx+n$$ but I don't get anything similar to the stuff in the passage above.

• Consider $$\begin{cases}x=x_0 + t\cos a\\y=y_0 + t \sin a.\end{cases}$$What happens if you determine $t$ from, say, the first equation and then plug it in the second one? – dfnu Nov 29 '19 at 20:50
• Equation of a line ? But how do I go the other way around ? – user728535 Nov 29 '19 at 21:04
• Let $m=\tan a$... – dfnu Nov 29 '19 at 21:05
• Still not clear. Let's sat that I want limit as $(x,y)\rightarrow (0,0)$ along the line $y=mx$. I transform all points that satisfy the line equation in polar form and it should be $lim_{x\to 0}f(x,mx) = lim_{t\to 0}f(tcosa, mtcosa)$, right ? Can you please write the full explanation in an answer ? How is limit along the line $l$ the same as it's written in passage above ? Maybe it has something to do with my pretty limited knowledge of analytical geometry. Thanks – user728535 Nov 29 '19 at 23:33

## 1 Answer

As you request I try to give more details in an answer. Suppose $$(x_0,y_0)\equiv (0,0)$$. You are calculating then the limit

$$\lim_{(x,mx)\to(0,0)} f(x,y).$$

Therefore, as $$x$$ goes to $$0$$, so does $$y$$, following the line $$r: y=mx.$$

Suppose the line forms an angle $$\alpha$$ with the positive $$x$$-semiaxis. Then $$r$$ can be expressed in parametric form as $$\begin{cases} x=t\cos\alpha\\ y=t \sin\alpha, \end{cases}$$ where $$t$$ runs through all real values, letting $$(x,y)$$ cover the entire line (with $$t=0$$ corresponding to the origin of the axes). Now your limit can be expressed as $$\lim_{t\to 0} f(t\cos\alpha,t\sin\alpha).$$

• I was not familiar with parametric form. Thanks. – user728535 Nov 30 '19 at 18:16
• @fridrih no problem, hope it did help. – dfnu Nov 30 '19 at 19:01
• One more question. Is there any real difference between this type of equation of a line and this: $$x = x_0 + tA, y= y_0 + tB$$ ? – user728535 Nov 30 '19 at 19:03
• @fridrih with generic $A$ and $B$ instead of $\cos \theta$ and $\sin \theta$, you mean? The only difference is that you're using a generic vector instead of a unit vector. If you divide your vector by $\sqrt{A^2+B^2}$ you get back to the sine/cosine version. – dfnu Nov 30 '19 at 19:07