1

I have a hard time evaluating $$ \lim_{(x, y)\to(0, 0)} \frac{e^{xy}\sin(xy)}{xy} $$

I keep getting the answer that the limit doesn't exist but actually the answer is $1$. Help.

Henrik supports the community
  • 6,831
Math
  • 437
  • 4
  • 10
  • 2
    How do you know what the answer is supposed to be? And why don't you show us your work, so we can point out the error? – Henrik supports the community Jul 05 '16 at 21:34
  • Are you able to determine the similar limit $\lim_{x \to 0} \frac{e^x \sin x}{x}$? – Brian Tung Jul 05 '16 at 21:37
  • Make the change of variable $r=xy$. Then use that $r\to 0$ if and only if $(x,y)\to (0,0)$. To solve for $r$ just use the L'Hopitals rule or the fact that $\sin(r)\sim r$ when $r\sim 0$. –  Jul 05 '16 at 21:37
  • 1
    @Jonh: Is "only if" true or necessary? – Brian Tung Jul 05 '16 at 21:38
  • @BrianTung I didn't meant to write it as some implication. I was trying to say that it os equivalent. Either way it is true, although the only if part might not be necessary –  Jul 05 '16 at 21:46
  • 1
    @Jonh It is not true that $xy \to 0$ implies $(x,y) \to (0,0)$. To see this, let $x$ be any fixed nonzero number and let $y \to 0$. Then $xy \to 0$ but $(x,y) \not\to (0,0)$. Fortunately for this problem, we only need the reverse implication, $(x,y) \to (0,0)$ implies $xy\to 0$, which is true (see my answer below). –  Jul 05 '16 at 22:10
  • @Bungo Absolutely, sorry, hurry-silly mistake –  Jul 05 '16 at 22:35

4 Answers4

2

Note that $(x,y) \to 0$ if and only if $x^2 + y^2 \to 0$. This fact combined with the inequality $|xy| \leq (x^2 + y^2) / 2$ means that the following chain of implications holds: $$(x,y) \to 0 \implies x^2 + y^2 \to 0 \implies xy \to 0$$ This justifies rewriting the problem by setting $u = xy$ and taking the limit as $u \to 0$: $$\lim_{u \to 0}\frac{e^u \sin(u)}{u} = \left(\lim_{u \to 0}e^u\right) \left( \lim_{u \to 0}\frac{\sin(u)}{u}\right) = (1)(1) = 1$$

2

Recall from elementary geometry that the sine function satisfies the inequalities

$$x\cos(x)\le \sin(x)\le x \tag 1$$

for $|x|\le \pi/2$. Then, we have from $(1)$

$$\cos(xy)\le \frac{\sin(xy)}{xy}\le 1 \tag2$$

In addition, I showed in This Answer using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$1+x\le e^x\le \frac{1}{1-x}$$

for $x<1$

Putting $(2)$ and $(3)$ together, we have

$$(1+xy)\cos(xy) \le \frac{e^{xy}\sin(xy)}{xy}\le \frac{1}{1-xy}$$

whereupon application of the squeeze theorem reveals

$$\bbox[5px,border:2px solid #C0A000]{\lim_{(x,y)\to (0,0)}\frac{e^{xy}\sin(xy)}{xy}=1}$$

Community
  • 1
Mark Viola
  • 179,405
1

Simply use the Taylor series $$e^{xy} = 1 + xy + \frac{x^2y^2}{2} + \frac{x^3 y^3}{6} +\cdots$$ and $$\sin(xy) = xy - \frac{x^3y^3}{6} + \frac{x^5y^5}{120} \pm \cdots$$ and you'll end up with the representation $$\frac{e^{xy}\sin(xy)}{xy} = 1 + \mathcal O(xy)$$ which holds as $(x,y) \to (0,0)$.

User8128
  • 15,485
  • 1
  • 18
  • 31
0

You can try polar coordinates for any point where the function is defined.:

$$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\implies \lim_{(x,y)\to(0,0)}e^{xy}\frac{\sin xy}{xy}=\lim_{r\to0}e^{r^2\cos t\sin t}\frac{\sin r^2cos t\sin t}{r^2\cos t\sin t}=e^0\cdot1=1$$