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I have to do the following limit:

$$\lim_{(x,y)\rightarrow (0,0)}\frac{e^{xy}}{x+1}$$

I did that if $x = 0 \Rightarrow \lim \rightarrow 1$ but if $y = 0 \Rightarrow \lim \rightarrow$ undefined. By taking $x=y$ it gets me to the same result. Is the limit then $1$ or in what other way can I evualte it to see if it exists and if so, what's its value?

Thanks for the help.

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    $\begingroup$ Why do you say the limit is undefined when $y=0$? $\endgroup$ – J. W. Tanner Jun 2 at 22:19
  • $\begingroup$ Sorry, I saw my mistake. Indeed with $y = 0$ the limit is still $0$ $\endgroup$ – Rolando González Jun 2 at 22:28
  • $\begingroup$ If the limits in the numerator and denominator both exist (denominator not zero), then you can just divide. $\endgroup$ – GEdgar Jun 2 at 22:59
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Actually, if $y=0$, it's $\lim\limits_{x\to0}\dfrac1{x+1}=\color{red}1$ too.

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In your class, certainly they should have mentioned that if you can 'plug it in' and the function is 'nice' (for an introductory level, essentially not piecewise), then the value is the limit. This is because at an introductory level, the functions we give you are (nearly) continuous functions. So the limit of the function will be the value at the limit (if it is defined). Notice $$ \lim_{(x,y) \to (0,0)} \dfrac{e^{xy}}{x+1} \stackrel{?}{=} \dfrac{e^0}{0+1}= \dfrac{1}{1}=1 $$ Therefore, the limit is actually exactly $1$. Again this uses the fact that the function is continuous. Notice that $e^{xy}$ is continuous for all $x,y$. [It's just like $e^z$, and you already knew that the function $e^x$ was continuous for any input $x$.] And the function $\dfrac{1}{x+1}$ is continuous for all $x \neq -1$ (notice you limiting point is $(0,0)$ so this isn't a problem). Then the product $e^{xy} \cdot \dfrac{1}{x+1}= \dfrac{e^{xy}}{x+1}$ is continuous for all $(x,y)$ with $x \neq -1$. But then the value of its limit is the just the value of the function at the point (assuming it is defined).

As another example(s), $$ \lim_{(x,y)\to(1,2)} \dfrac{x+y}{x-y}= \dfrac{1+2}{1-2}= \dfrac{3}{-1}= -3 $$ $$ \lim_{(x,y) \to (\pi,3)} \dfrac{\sin y \cos x}{\ln(x+y)}= \dfrac{\sin 3 \cos \pi}{\ln(3+\pi)}= \dfrac{-\sin 3}{\ln(3+\pi)} $$ Again where this breaks down is if you have a piecewise function, for instance $$ f(x,y)= \begin{cases} \sin(x+y), &(x,y) \neq (0,0) \\ 5,& (x,y)= (0,0) \end{cases} $$ Then the limit $\lim_{(x,y) \to (0,0)} f(x,y)$ might not be so obvious. This also breaks down if the function is undefined at the point. For example, $$ \lim_{(x,y) \to (0,0)} \dfrac{x^2-y^2}{x+y} $$ where plugging in will get $0/0$. [Notice there you need the 'trick' of factoring before you can simply 'plug in' and find the limit of $0$.]

Note. For more knowledgeable readers, of course, continuity comes after limits - how else would you define it. But for more elementary multivariable calculus courses, it often suffices to retroactively justify the 'plug 'n chug method' later when discussing continuity once students have their head around limits, just as in the single variable case in Calculus I.

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