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I am looking at : $$\lim\limits_ {(x, y) \to (0, 0)} \frac {e^{xy} − 1} y$$

and choosing the path $y = 0$, the limit becomes:

$$\lim\limits_ {(x, 0) \to (0, 0)} \frac {1 − 1} 0$$

On an answer key, this limit evaluates to 0 but I don't know why because 0/0 should be undefined.

Guy Fsone
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fdfdsa
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    You're actually correct that the limit isn't defined until you (they) define the function $f(x,y)$ when $y=0$. However, what is intended is to realize that $e^{xy}-1 \approx xy$, so the function looks like $xy/y = x$ for small $(x,y)$. – Ted Shifrin Jan 27 '18 at 01:52
  • Welcome to MSE. Please use MathJax. – José Carlos Santos Jan 27 '18 at 01:55
  • @TedShifrin How would one show that $e^{xy}-1\approx xy$ for small $x,y$? – actinidia Jan 27 '18 at 02:23
  • @TiwaAina What's the derivative of $f(x)=e^x$ at $x=0$? – Ted Shifrin Jan 27 '18 at 03:23
  • @TedShifrin 1, and subtracting 1 from that gives zero, which is about equal to the product of two small numbers x,y? Is that where you'd go with that observation? – actinidia Jan 27 '18 at 03:27
  • @TiwaAina No. write out the linear approximation formula (equation of the tangent line). – Ted Shifrin Jan 27 '18 at 03:29
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    @TiwaAina.[1}. Define $\log x=\int+1^x\frac {1}{z}dz.$ Note $\frac {d\log x}{dx}=\frac {1}{x}.$ ...[2]. $\log (xy)-\log x=\int_x^{xy}\frac {1}{z}dz.$ Let $z=z'x.$ Then $\log (xy)-\log x=\int_1^y\frac {1}{z'}dz'=\log y.$ Therefore the inverse function $\log^{-1}(x)=e^x$ for all $x$, where $e$ satisfies $\int_1^e\frac {1}{z}dz=1.$...[3] With $w=e^x$ we have $dw/dx=(dx/dw)^{-1}=((d\log w)/dw)^{-1}=(1/w)^{-1}=w=e^x.$ Therefore, with $w'=dw/dx$ we have $1=e^0=w'(0)=\lim_{z\to 0}\frac {e^z-e^0}{z-0}=\lim_{z\to 0}\frac {e^z-1}{z}. $ – DanielWainfleet Jan 27 '18 at 03:30
  • @TedShifrin $L(x) = f(0,0) + 1(x-0)+1(y-0) = x + y$. If I did that correctly, then we'd need to show that x+y = xy for small x,y which I don't think is true. – actinidia Jan 27 '18 at 03:38
  • @TiwaAina. That should say: [1]. Define $\log x = \int_1^x \frac {1}{z}dz.$ – DanielWainfleet Jan 27 '18 at 03:38
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    @TiwaAina: No. If you're going to use multivariable calculus, then you have to go to the second-order Taylor polynomial, as the first partials vanish at the origin. I was suggesting using $e^u \approx 1 + u$ and then substituting $u=xy$. – Ted Shifrin Jan 27 '18 at 03:41

5 Answers5

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The path ${y=0}$ is not in the domain of your function.

For $y\ne 0$, we have $$e^{xy}=1+xy+\frac {x^2y^2}{2!}+.....$$

$$ e^{xy}-1 = xy+\frac {x^2y^2}{2!}+.....$$

$$ \frac{e^{xy}-1}{y} = x+\frac {x^2y}{2!}+.....$$

$$ lim_{(x,y)\to (0,0)} \frac{e^{xy}-1}{y} =$$

$$ lim_{(x,y)\to (0,0)} x+\frac {x^2y}{2!}+..... = {0} $$

Mohammad Riazi-Kermani
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I thought it might be instructive to present an approach that relies only on a pair of elementary inequalities and the squeeze theorem. To that end, we proceed.

In THIS ANSWER, I used only the limit definition of the exponential function and Bernoulli's Inequality to show that

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

for $x<1$. Using $(1)$, we have for $xy<1$

$$x\le \frac{e^{xy}-1}{y}\le \frac{x}{1-xy}$$

whence the squeeze theorem guarantees that

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

And we are done!

Mark Viola
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  • Please let me know how I can improve my answer. I really want to give you the best answer I can. I'm happy to delete the answer if you did not find it useful. – Mark Viola Mar 07 '18 at 00:45
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To evaluate a limit $f(z)/g(z)$ as $z\to 0$ you generally cannot just take $f(0)/g(0).$ For example if $f(z)=g(z)=z$ for every $z,$ then $f(z)/g(z)=1$ when $z\ne 0$ but $f(0)/g(0)$ does not exist.

It is tacitly assumed that if $g(0)=0$ then the limit of $f(z)/g(z)$ as $z\to 0 $ is evaluated as the limit as $z\to 0$ through non-zero values.

(i). For $y\ne 0\ne x$ we have $$\frac {e^{xy}-1}{y}=x\cdot \frac {e^{xy}-1}{xy}=x\cdot \frac {f(z)-f(0)}{z-0}$$ where $z=xy$ and $f(z)=e^z.$ As $(x,y)\to (0,0)$ with $y\ne 0\ne x$ we have $z\to 0$ so $$\frac {f(z)-f(0)}{z-0}\to f'(0)=e^0=1.$$ And we also have $x\to 0,$ so $x\cdot \frac {f(z)-f(0)}{z-0}\to 0\cdot 1=0.$

(ii). For $y\ne 0$ and $x=0$ we have $\frac {e^{xy}-1}{y}=\frac {e^0-1}{y}=0.$

By (i) and (ii), if $(x,y)\to (0,0)$ with $y\ne 0$ then $\frac{e^{xy}-1}{y}\to 0.$

DanielWainfleet
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$$\lim\limits_ {(x, y) \to (0, 0)} \frac {e^{xy} − 1} y= \lim\limits_ {(x, y) \to (0, 0)} x\cdot\lim\limits_ {(x, y) \to (0, 0)}\frac {e^{xy} − 1}{ xy}= \lim\limits_ {{(x, y) \to (0, 0)}} x\cdot\lim\limits_ {h \to 0}\frac {e^{h} − 1}{ h}=0*1=0$$

Guy Fsone
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Note that

$$\frac {e^{xy} − 1} y=x \cdot \frac {e^{xy} − 1} {xy}\to 0\cdot1=0$$

indeed

$$e^{xy}=1+xy+o(\rho^2)\implies \frac {e^{xy} − 1} {xy}=\frac {1+xy+o(\rho^2) − 1} {xy}=\frac {xy+o(\rho^2)} {xy}=1+o(1)\to1$$

user
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  • I am very late to this, but how do you know that $o(\rho^2)/(xy)$ goes to $0$? – user210089 Nov 17 '23 at 13:42
  • @user210089 We have that $o(\rho^2)=o(x^2+y^2)=o(xy)$. – user Nov 17 '23 at 21:26
  • Hey, thanks for answering. I am not sure I am understanding correctly. Consider $f(x,y) = (x^2+y^2)x$. We have $\lim_{(x,y)\to (0,0)} \frac{f(x,y)}{x^2+y^2} = 0$, so $f\in o(x^2+y^2)$. However, $\lim_{(x,y)\to (0,0)} \frac{f(x,y)}{xy}=\lim_{(x,y)\to (0,0)} \frac{x^2+y^2}{y}$, which does not exist, so $f\notin o(xy)$. – user210089 Nov 17 '23 at 21:47
  • @user210089 We have that, by definition, $o(xy)=xy \cdot \omega(xy)$ with $\omega(xy)\to 0$ and $xy \cdot \omega(xy)=(x^2+y^2)\cdot \frac{xy}{x^2+y^2}\omega(xy)=(x^2+y^2)\cdot \omega_1(x,y)$ with $\omega_1(x,y) \to 0$ which means by definition $o(xy)=o(x^2+y^2)$. – user Nov 17 '23 at 22:40
  • You have proven that $o(xy)\subset o(x^2+y^2)$, using that $\frac{xy}{x^2+y^2}$ is a bounded function. But in the limit you are using $o(x^2+y^2)\subset o(xy)$. So you have some function $g(x,y) = (x^2+y^2)w(x,y)$, how do you express it as $xy \cdot w_1(x,y)$ with $w_1(x,y)\to 0$? – user210089 Nov 17 '23 at 23:03
  • In my example, $f(x,y)=(x^2+y^2)x \in o(x^2+y^2)$, since $f(x,y) = (x^2+y^2)w(x,y)$, being $w(x,y) = x\to 0$ as $(x,y)\to (0,0)$. However, $f\notin o(xy)$, since then $f(x,y) = xy\cdot w_1(x,y)$ with $w_1(x,y) \to 0$ as $(x,y)\to (0,0)$, so $\lim_{(x,y)\to (0,0)}\frac{f(x,y)}{xy} = \lim_{(x,y)\to (0,0)} w_1(x,y) = 0$, and I showed this is not the case. – user210089 Nov 17 '23 at 23:13
  • @user210089 Yes ok I see your point, note that here I'm using $o(\rho^2)$ instead of $o(xy)$ so that $o(xy)/xy=o(1)\to 0$. If you thonkit is confusing I can change the notation. – user Nov 17 '23 at 23:33
  • What I don't know how to prove is that the remainder $R(x,y)$ of the Taylor expansion of $exp(xy)$, which we know belongs to $o(x^2+y^2)$, verifies that $\lim_{(x,y)\to (0,0)} \frac{R(x,y)}{xy}=0$ (that is, belongs to $o(xy)$). Perharps I should post this as a question, and maybe you could expand your answer? – user210089 Nov 17 '23 at 23:41
  • Regarding the notation, you mean that you are using $o(\rho ^2)$ instead of $o(x^2+y^2)$? If so, its fine for me. – user210089 Nov 17 '23 at 23:42
  • @user210089 For my knowledge, in multivariable case using Peano's notation for the remainder, that is $f(x,y)=P_n(x,y)+r_n(x,y)$ we mean that $r_n(x,y)=o(\rho^n)$ that is $\frac{r_n(x,y)}{\rho^n}\to 0$. In our case $r_n(x,y)=\frac12x^2y^2+\frac16x^3y^3+\ldots$ and $\frac{r_n(x,y)}{x^2+y^2} \to 0$. – user Nov 18 '23 at 10:58
  • Let us continue this discussion in chat. – user210089 Nov 18 '23 at 11:08