Prove, with the help of the definition of convergence, that $\lim_{(x,y) \to (-1,1)} (3x^2 + 4y^2) = 7$. I need to solve the following problem:
Prove, with the help of the definition of convergence, that
$\lim_{(x,y) \to (-1,1)} (3x^2 + 4y^2) = 7$
The definition of convergence, as given in our textbook, goes like this:
Consider a two variable function $f(x,y)$. If for each $ε>0$ there is a $δ>0$ such as for each $(x,y)\in D(f)$ with $0<|x - a|<δ$ and $0<|y - b|<δ$ then $|f(x,y) - l|<ε$ and $\lim_{(x,y) \to (a,b)} f(x,y)=l$ .
I have a hard time understanding the definition itself, so I tried following the closest example in our  textbook, presented below:
Prove, with the help of the definition of convergence, that
$\lim_{(x,y) \to (2,1)} (x^2 - 3y) = 1$
Solution:
Considering $f(x,y) = x^2 -3y /R^2$ then:
$|f(x,y)-1| = |x^2-3y-1| = |(x^2 -2^2)-3(y-1)|\leq|x-2||x+2|+3|y-1|$
so if $δ>0$ with $|x-2|<δ$ and $|y-1|<δ$ then $|x+2|\leq|x-2|+4<δ+4$
and
$|f(x,y)-1|<δ(δ+4)+3δ=δ^2+7δ$ (Relation 1)
For $ε>0$, if $δ=min\{1,\frac{ε}{8}\}$ is taken, then, according to (Relation 1), it will be
$|f(x,y)-1|<δ^2+7δ<8δ<ε$
So we get that $\lim_{(x,y) \to (2,1)} (x^2 - 3y) = 1$
Trying to follow the example for $\lim_{(x,y) \to (-1,1)} (3x^2 + 4y^2) = 7$, this time, I came up with this:
Considering $f(x,y)=3x^2 + 4y^2/R^2$ then:
$|f(x,y)|=|3x^2 + 4y^2 -7|=|3(x^2-1)+4(y^2-1)|\leq3|x-1||x+1|+4|y-1||y+1|$
so for $δ>0$ with $|x+1|<δ$ and $|y-1|<δ$ we have that
$|x-1|\leq|x+1|+1<δ+1$ and $|y+1|\leq|y-1|+1<δ+1$
and
$|f(x,y)-7|<3δ(δ+1)+4δ(δ+1)=3δ^2+3δ+4δ^2+4δ<7δ^2+7δ$ (Relation 1)
For $ε>0$ and $δ=\frac{ε}{14}$, we have, according to (Relation 1) that
$|f(x,y)-7|<7δ^2+7δ<14δ<ε$
So, we get that $\lim_{(x,y) \to (-1,1)} (3x^2 + 4y^2) = 7$
Is this solution correct? If yes, can I apply something similar for $\lim_{(x,y) \to (2,-1)} (3xy+x^2-2y)=0$ ? Also, can someone elaborate more on the definition of convergence I gave?
Thank you in advance.
 A: You've made a couple of mistakes.  You say
"so for $\delta>0$ with $|x+1|<\delta$ and $|y−1|<\delta$ we have that $|x-1|\leq|x+1|+1<\delta+1$ and $|y+1|\leq|y-1|+1<\delta+1$," and this isn't so.
We have $$|x-1|=|x+1-2|\leq|x+1|+2<\delta+2$$
You've made the same error with $|y+1|$.
The definition of limit in the case of functions of several variables, means the same thing intuitively that it means in the case of functions of a single variable.  The function value is arbitrarily close to the limit (within $\varepsilon$) whenever it is evaluated at points sufficiently close to the given point (within $\delta$).  We can either require that each coordinate is within $\delta$ as in the definition above, or we can require that the distance from $(x,y)$ to $(x_0, y_0)$ is $<\delta$.  It come to the same thing, because if each coordinate is within $\delta$ the distance is $<\sqrt2\delta$.  Does this make it any clearer?
A: Since saulspatz' answer discusses the OP's work, I am free to present two alternative approaches, one easy and one harder.
The easy way is to use the theorem (adapted for two variables) that if $f$ is a continuous function in a neighborhood around $(-1,1),$ then the $\displaystyle \lim_{(x,y) \to (-1,1)} f(x,y) = f(-1,1).$
Obviously, this isn't the approach that the OP wants, but it is still worth mentioning.

My $\epsilon, \delta$ approach:
I want $|f(x,y) - f(-1,1)| < \epsilon$. 
This means that I want
$-\epsilon < [(3x^2 + 4y^2) - 7] < \epsilon.$ 
With this in mind, I will now focus on $\delta$.
I will derive a candidate specification for $\delta$ in terms of $\epsilon$, and then verify that the candidate specification works.
$0 < |(x,y) - (-1,1)| < \delta \implies$ 

*

*$(x,y) \neq (-1,1)$.


*$0 \leq |x - (-1)| < \delta.$


*$0 \leq |y - (1)| < \delta$.
The above conclusions are based on the idea that if $(x,y)$ is any point inside the circle of radius $\delta$, centered at the origin, then $|x| < \delta$ and $|y| < \delta$.
So, now I have that :

*

*$-\delta < (x + 1) < \delta \implies -\delta - 1 < x < \delta - 1.$


*$-\delta < (y - 1) < \delta \implies -\delta + 1 < x < \delta + 1.$
I always try to simplify everything to linear constraints.  In the present problem, I will employ the artificial contrivance of adding the constraint that $\delta < 1$.  This will automatically imply that $\delta^2 < \delta$.
At this point, I have that 
$(-\delta -1) < x < (\delta - 1) < 0.$ 
Therefore, 
$(\delta - 1)^2 < x^2 < (-\delta - 1)^2.$ 
Therefore, $\delta^2 - 2\delta + 1 < x^2 < \delta^2 + 2\delta + 1.$
Since $\delta < 1 \implies \delta^2 < \delta$, 
I can presume that 
$1 -3\delta < x^2 < 1 + 3\delta.$
Note that if I wanted, I could certainly tighten the bounds.  I much prefer bounds that are very simple.
Similarly, 
$0 < (-\delta + 1) < y < (\delta + 1) < 0.$ 
Therefore, 
$(1 - \delta)^2 < y^2 < (1 + \delta)^2.$ 
Therefore, $\delta^2 - 2\delta + 1 < y^2 < \delta^2 + 2\delta + 1.$
Therefore, 
$1 -3\delta < y^2 < 1 + 3\delta.$
Therefore,
$7 - 21\delta < 3x^2 + 4y^2 < 7 + 21\delta$.
Therefore,
$-21\delta < (3x^2 + 4y^2 - 7) < 21\delta$.
Remembering my goal of 
$-\epsilon < [(3x^2 + 4y^2) - 7] < \epsilon$ 
I need both $\delta < 1$ and $21\delta < \epsilon.$
Therefore, I will choose a candidate specification of
$$\delta = \min \left(\frac{\epsilon}{25}, \frac{1}{2}\right).$$
Verification:
Now, step by step, I need to verify that the candidate specification works.
I can assume, from the specification that 
$21 \delta < \epsilon$ and that $\delta < 1.$
Also, I am starting from the premise that 
$|(x,y) - (-1,1)| < \delta.$
Therefore, as previously analyzed, I know that 
$(-1 - \delta) < x < (\delta - 1) < 0.$ 
Therefore, as previously analyzed, I know that 
$1 - 3\delta < x^2 < 1 + 3\delta.$
Repeating the analysis for $y$,
I also have that 
$1 - 3\delta < y^2 < 1 + 3\delta.$
Therefore, $7 - 21\delta < 3x^2 + 4y^2 = f(x,y) < 7 + 21\delta.$
Therefore, $-21\delta < (3x^2 + 4y^2 - 7) = f(x,y) - f(-1,1) < 21\delta.$
Since $21\delta < \epsilon$, this implies that 
$-\epsilon < f(x,y) - f(-1,1) < \epsilon$ as required.

The above approach begs the question: since the verification is nothing but a re-hash of the original analysis, what is the point of the verification.
The derivation started with the premise that 
$-\epsilon < f(x,y) - f(-1,1) < \epsilon$ 
and implied from this that 
$\delta < 1$ and $21\delta < \epsilon$.
The direction of these implications is backwards from what is needed.  The implication to be demonstrated is that 
$0 < |(x,y) - (-1,1)| < \delta \implies |f(x,y) - f(-1,1)| < \epsilon$.
Further, sometimes, when you take a derivation step, the implication won't actually be a two-way implication.  For example, 
$x = 2 \implies x^2 = 4$, but the reverse implication is false.
During the derivation process, I intentionally avoided even considering whether any implications were only one-way.  Therefore, I re-hashed the analysis by verifying the candidate specification.
