# Proving that $\lim_{(x,y)\to(0,2)} |y|^x(x+1)^y = 1$ using the limit definition

I started with this strategy:

$$||y|^x(x+1)^y - 1| = \bigg||y|^x \bigg((x+1)^y -\frac{1}{|y|^{x-1}} + \frac{1}{|y|^x} \bigg) + |y| - 2 \bigg| \leq \bigg||y|^x \bigg((x+1)^y -\frac{1}{|y|^{x-1}} + \frac{1}{|y|^x} \bigg) \bigg| + ||y| - 2 |$$

But it soon went awry. Should I use $$e^{\ln x} = x$$ and/or $$x-1 \leq e^{x} - 2$$? I'm completely lost.

What we have here is a product of two similar-looking exponential functions. I will prove

$$\lim_{(x,y) \rightarrow (0,2)} |y|^{x} = 1.$$

Let $$\epsilon > 0$$ be arbitrary. We seek a $$\delta > 0$$ such that $$|(x,y-2)| < \delta$$ implies $$||y|^{x} - 1| < \epsilon$$. No matter which $$\delta$$ we select, the following is true:

$$||y|^{x} - 1| = \begin{cases} 1 - |y|^{x} \leq 1 - (2+\delta)^{-\delta}, \ \ \ |y|^{x} - 1 \leq 0 \\ |y|^{x} - 1 \leq (2+\delta)^{\delta} - 1, \ \ \ |y|^{x} - 1 \geq 0, \end{cases}$$

where we use the reverse triangle inequality to get $$|y| \leq 2+\delta$$. In the first case above, we know $$x \geq -\delta$$.

Focus on the second case: We want to select $$\delta$$ such that $$\epsilon \geq (2+\delta)^{\delta} - 1$$. An equivalent condition is $$\ln(\epsilon+1) \geq \delta \ln(2+\delta)$$. A sufficient condition is $$\ln(\epsilon + 1) \geq \delta(2+\delta)$$, since $$\ln(z) \leq z$$ for any $$z > 0$$. By solving this inequality for $$\delta$$, we get an even more sufficient condition of $$\delta \leq \sqrt{\ln(\epsilon + 1) + 1} - 1$$. By selecting $$\delta$$ in this way, we will get $$|y|^{x} - 1 < \epsilon$$.

First case: We have $$1 - (2+\delta)^{-\delta} = 1 - e^{-\delta\ln(2+\delta)} \leq \delta\ln(2+\delta)$$ by the property $$1 - e^{-z} \leq z$$ for any real $$z$$. If we assume the condition for $$\delta$$ in the second case, then $$\delta \ln(2+\delta) < \ln(\epsilon+1) \leq \epsilon$$ (this last inequality is another property of $$\ln(\cdot)$$).

This shows that selecting $$\delta = \sqrt{\ln(\epsilon+1)+1} - 1$$ works for both cases of $$||y|^{x} - 1|$$ being less than $$\epsilon$$.

Showing $$\lim_{(x,y) \rightarrow (0,2)}(x+1)^{y} = 1$$ has almost identical steps, so I will let you do that. Once you've done this, you will get a second choice for $$\delta$$. If you pick $$\delta$$ to be the minimum of these two choices, then your limit statement will hold. Your analysis textbook probably has a proof for the product of two functions, so you might be able to just cite that.