Limits of $ f(x,y) = y\ (1-x)^{y-2} $ reach contradiction

Question

Introduction

I evaluated a limit of a multivariable function at two values and combining the two results, gives a contradiction. I want to understand where my mistake is: the limit evaluations or the interpretation of the results.

Results

The function is $f:R^2 \to R$ defined by $$ f(x,y) = y\ (1-x)^{y-2} $$

I now list the two limits I evaluated:

$$\lim_{x \to 0^{+} \\ y \to \infty}{f(x,y)} = \infty$$ $$\lim_{x \to x_0 \in (0,1) \\ y \to \infty}{f(x,y)} = 0$$

Remark: I proved both results with $\epsilon,\delta$ notation. If needed I can type up the proofs and include them in the question

Intuition of the results

Intuitively I initially was ok with what I showed above. I thought that $f(x,y)$ simply makes a bigger and bigger jump from $f(\delta,y)$ to $f(0,y)$ as $y$ becomes larger and larger. However when I looked closer at the results I arrived at a contradiction as I will demonstrate below. Checking with some 3D plotter, simply reinforces this discussion, indicating that the limit values are correct.

The Contradiction

Now, to produce this contradiction first I had to translate the two limits above into their definition.

$$\lim_{x \to 0^{+} \\ y \to \infty}{f(x,y)} = \infty \Rightarrow$$ $$ \begin{equation} \forall N>0 \ , \ \exists \ \delta,M > 0 \text{ such that } f(x,y)>N \\ \forall x \in (0,\delta) \text{ and } y \in (M,\infty) \end{equation} $$

and for the second one:

$$\lim_{x \to x_0 \in (0,1) \\ y \to \infty}{f(x,y)} = 0\Rightarrow$$ $$ \begin{equation} \forall \epsilon^{\prime}>0 \ , \ \exists \ \delta^{\prime},M^{\prime} > 0 \text{ such that } |f(x,y)|<\epsilon^{\prime} \\ \forall |x-x_0| < \delta^{\prime} \text{ and } y \in (M^{\prime},\infty) \end{equation} $$ So now for the contradiction:

Step 1: Pick some $N>0$. By the first limit, $$\exists \ \delta,M > 0 \text{ such that} f(x,y)>N \\ \forall x \in (0,\delta) \text{ and } y \in (M,\infty)$$

Step 2: Pick some $x_0 \in (0,\delta) \text{ and let } \epsilon^{\prime} = N$. By the second limit we get that: $$\exists \ \delta^{\prime},M^{\prime} > 0 \text{ such that} |f(x,y)|<N \\ \forall |x-x_0| < \delta^{\prime} \text{ and } y \in (M^{\prime},\infty)$$

Step 3: Pick some $x^* \in (x_0,\min{(\delta,x_0 + \delta^{\prime})})$ and $y^* \in (\max(M,M^{\prime}),\infty)$. It is evident that there exists such $x^*$, $y^*$ and they both satisfy both of the limit definitions above.

Step 4: By using the two limit definitions, we get that: $f(x^*,y^*) > N$ and $f(x^*,y^*) < N$ since $f>0 \ \forall \ x<1 \text{ and } y>2$, which means that $f(x^*,y^*) > f(x^*,y^*)$, a contradiction.

Conclusion

My best guess is that I have somehow messed up the definitions of the limits when one variable tends to infinity.

What I am asking is: Which false assumption am I making?

Answer 1

One can see that if $(x,y)$ tends to $(0,\infty)$ through the curve $y = 1/x^2$ the limit of $f(x,y)$ is $0$. On the other hand, if $(x,y)$ tends to $(0,\infty)$ through the curve $y = 1/x$ the limit of $f(x,y)$ is $\infty$.

This proves that $\displaystyle\lim_{x\to 0^+ \\ y\to\infty}f(x,y)$ doesn't exists.