Calculus limit with 2 variables

Question

How can i proof that the limit does not exist:

$\lim_{(x,y)\to (1,1)} \frac{xy^2 - 1}{y-1}$

Thanks in advance!

Answer 1

Fix $x=1$ and take $y\to 1$ and you should get one limit. Now take $y=x$ and take the limit as $y\to 1$. Are they the same?

Answer 2

To show that the limit does not exist take two different pairs of sequences with limit $1$, and show that the limit in both cases is not equal.

If we approach $x$ and $y$ similar, we can set $x=y$ for starters. Then we get: $$ \lim_{y\to 1} \frac{y^3-1}{y-1}=\lim_{y\to 1}(y^2+y+1)=3 $$

Setting $x=1$ however we get $$ \lim_{y\to 1} \frac{y^2-1}{y-1}=\lim_{y\to 1}(y+1)=2 $$

So the limit does not exist.

Answer 3

Since $\Bbb R^n$ is first countable we can use the following theorem

Theorem

If $X$ is first countable and if $f:X\to Y$ is a function then $y_0$ is the limit of $f$ as $x$ approaches at $x_0$ if and only if for any sequence $(x_n)_{n\in\Bbb N}$ converging to $x_0$ it happens that $\left(f(x_n)\right)_{n\in\Bbb N}$ converges to $y_0$.

Proof. See here.

So if we put $f(x,y):=\frac{xy^2-1}{y-1}$ then we consider the sequences $a_n:=\Big(\frac{1}n+1,\frac{1}n+1\Big)$ and $b_n:=\Big(\frac{1}{n^3}+1,\frac{1}n+1\Big)$ converging to $(1,1)$ and we observe that

$$ \lim_{n\rightarrow\infty}f(a_n)=3\neq2=\lim_{n\rightarrow\infty}f(b_n) $$ so that your statement follows directely form the above theorem.