Does the limit exist ? and how to compute it?

Question

$$\displaystyle\lim_{(x,y)\rightarrow (1,0)}\frac{x-1}{\sqrt{(x-1)^2+y^2}}$$

By direct substitution that's a ( $ \frac{0}{0}$ ) undefined

so can I approach it by polar equation or by different paths ?

Mine was the polar form considering eqaution of circle is :

$$a(x-x_1)^2+b(y-y_1)^2$$

Answer 1

Different paths are useful in this exercise, easy to choose and compute.

If $x=1$ you get $0.$

If $x>1$ and $y=0,$ the result is $1.$

This is enough to conclude that the limite does not exist.

Answer 2

Let $x=u+1, \, v=y$ with $(u,v)\to (0,0)$ then

$$\lim_{(x,y)\rightarrow (1,0)}\frac{x-1}{\sqrt{(x-1)^2+y^2}}=\lim_{(u,v)\rightarrow (0,0)}\frac{u}{\sqrt{u^2+v^2}}$$

and since by polar coordinates $\frac{u}{\sqrt{u^2+v^2}}=\cos \theta$ we can conclude that the limit doesn't exist, indeed

• $u=0 \implies \frac{u}{\sqrt{u^2+v^2}}=0$

• $u=v \implies \frac{u}{\sqrt{u^2+v^2}}=\frac{\sqrt 2}2$

Answer 3

Put

$$x=1+r\cos(t)$$ and

$$y=r\sin(t)$$

the function becomes

$$\frac{r\cos(t)}{r}=\cos(t)$$

the limit depends on the angle $t$, so The limit doesn't exist.