1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.