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.


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