prove that:
$\displaystyle lim_{(x,y)\to (1,1)}\frac{2+x}{y} = 3$
Again i know the definition but i have no idea how to get δ
Are you required to use and "$\displaystyle \epsilon, \delta$" proof?
You can, for example, replace x with x'= x- 1 so x= x'+1 and y with y'= y- 1so y= y'+1. Then, the problem becomes to prove that $\displaystyle \frac{2+ x'+1}{y'+1}= \frac{3+ x'}{y'+1}$ goes to 3 as x' and y' go to 0. Now change to polar coordinates: $\displaystyle x'= r cos(\theta)$ and $\displaystyle y'= r sin(\theta)$. Now we have $\displaystyle \frac{r cos(\theta)+ 3}{r sin(\theta)+ 1}$. The reason I "shifted" the point to (0, 0) is that r alone measures the distance to the origin. Now, as r goes to 0, both $\displaystyle r cos(\theta)$ and $\displaystyle r sin(\theta)$ go to 0 as r goes to 0, no matter what $\displaystyle \theta$ is and so the limit is 3.
And, now, I think it would not be too difficult to get a $\displaystyle \delta, \epsilon$ proof from that. Given any epsilon, we want $\displaystyle \delta$ so that if $\displaystyle \sqrt{(x-1)^2+ (y-1)^2}= r< \delta$ then $\displaystyle |\frac{2+ x}{y}- 3|= |\frac{rcos(\theta)+ 3}{r sin(\theta)+ 1}- 3|$$\displaystyle = \frac{r cos(\theta)- 3r sin(\theta)}{rsin(\theta)+ 1}|= r|\frac{cos(\theta)- 3sin(\theta)}{r sin(\theta)+ 1}|< \epsilon$
Find an upper bound on $\displaystyle |\frac{cos(\theta)- 3sin(\theta)}{r sin(\theta)+ 1}|$ and take $\displaystyle \delta$ to be less than $\displaystyle \epsilon$ divided by that upper bound.
Why ,do we have to make so many substitutions ,can we not in straight forward manner using the ε-δ definition find the limit?
The definition i was given is:
given ε>0 find a δ>0 such that :
if $\displaystyle 0<||(x,y)-(1,1)||_{Eu}<\delta$ ,then $\displaystyle |\frac{2+x}{y}-3|<\epsilon$