I have tried many different methods, I can't find any upper bound for squeeze theorem. Polar coordinates create big, messy limit which is hard to evaluate. WolframAlpha suggests that the limit doesn't exist, but on the graph it looks like the limit is $0$. I tried finding counterexample but all lines going through $(1,1)$ produce $0$ as the limit.
$\begingroup$
$\endgroup$
2
-
2$\begingroup$ Actually the limit exists and it's $0$. Don't trust Wolfram Alpha, it's trash. $\endgroup$– Enrico M.Mar 29, 2020 at 18:24
-
$\begingroup$ Ok, but how can I show that it is indeed $0$? $\endgroup$– kalashotMar 29, 2020 at 18:29
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
6
For $y<1$ we obtain: $$x^y=1+y(x-1)-\frac{(1-y)y(x-1)^2}{2}+\frac{(2-y)(1-y)y(x-1)^3}{6}-\frac{(3-y)(2-y)(1-y)y(\theta x)^4}{24}\leq$$ $$\leq1+y(x-1)-\frac{(1-y)y(x-1)^2}{2}+\frac{(2-y)(1-y)y(x-1)^3}{6}.$$ Thus, $$\frac{x^y-y(x-1)-1}{(x-1)^2+(y-1)^2}\leq\frac{(x-1)^2}{(x-1)^2+(y-1)^2}\left(-\frac{y(1-y)}{2}+\frac{y(1-y)(2-y)(x-1)}{6}\right)\rightarrow0.$$ Can you end it now?
answered Mar 29, 2020 at 18:33
-
$\begingroup$ I wonder if we can use Taylor's theorem in multivariable functions? $\endgroup$– kalashotMar 29, 2020 at 18:42
-
$\begingroup$ @dominikstepien2000 Firstly, we can. But in our case $y$ is a constant. $\endgroup$ Mar 29, 2020 at 18:44
-
$\begingroup$ Why $y$ is a constant if afterwards we take a limit in terms of $x$ and $y$? $\endgroup$– kalashotMar 29, 2020 at 18:47
-
$\begingroup$ @dominikstepien2000 We can get an estimations for $x^y$ around the point $(1,1)$ After this work we can get that the limit is equal to $0$. $\endgroup$ Mar 29, 2020 at 18:49
-
$\begingroup$ @Micheal Rozenberg Ok, but I have few more questions isn't the right side of last inequality less then zero when $y$ is close to 1? And my second question is of the same manner, is the second inequality true if $y>1$? $\endgroup$– kalashotMar 29, 2020 at 18:54
$\begingroup$
$\endgroup$
2
On the line $y = x$ your function becomes
$$f(x, x) = \frac{x^x-(x-1) x-1}{2 (x-1)^2}$$
And it's rather easy to check that $\lim_{x\to 1} f(x, x) = 0$.
On the line $y = -x$ the function becomes
$$f(x, -x) = \frac{x^{-x}+(x-1) x-1}{2 \left(x^2+1\right)}$$
And again $\lim_{x\to 1} f(x, -x) = 0$
You can go even on the lines $y = \alpha x$ for $\alpha \in \mathbb{R}$ and you will get the same result. Indeed:
$$f(x, \alpha x) = \frac{x^{a x}-a (x-1) x-1}{(a x-1)^2+(x-1)^2}$$
And again $\lim_{x\to 1} f(x, \alpha x) = 0$
answered Mar 29, 2020 at 18:38
-
$\begingroup$ It only shows that limit is 0 on those lines and not in general. $\endgroup$– kalashotMar 29, 2020 at 18:41
-
$\begingroup$ @dominikstepien2000 The use Taylor series! Or check all the nonlinear lines :) $\endgroup$ Mar 29, 2020 at 18:51