# Multivariable Limits

Can someone help me calculate the following limits?

1) $\displaystyle\lim _ {x \to 0 , y \to 0 } \frac{\sin(xy)}{\sqrt{x^2+y^2}}$ (it should equal zero, but I can't figure out how to compute it ) .

2) $\displaystyle\lim_ {(x,y)\to (0,\frac{\pi}{2} )} (1-\cos(x+y) ) ^{\tan(x+y)}$ (it should equal $1/e$).

3) $\displaystyle\lim_{(x,y) \to (0,0) } \frac{x^2 y }{x^2 + y^4 }$ (which should equal zero).

4) $\displaystyle \lim_{(x,y) \to (0,1) } (1+3x^2 y )^ \frac{1}{x^2 (1+y) }$ (which should equal $e^{3/2}$ ).

Any help would be great !

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Compute limit of variable x then variable y if it is the same as limit of y then x , you have a limit other wise it is meaningless limit. –  Arjang Jul 9 '12 at 6:12
What you're actually saying is that if I first take the limit on x and then on y, it should give me the same result as taking them both together to the limit? –  joshua Jul 9 '12 at 6:13
Well, it actually helps me in the first and third parts. But how can I know that the double limit exists in the first place? –  joshua Jul 9 '12 at 6:16
the double limit exists if the limit is not order dependent, that is the limit of x first then y is the same as limit of y first then x. –  Arjang Jul 9 '12 at 6:18
@Arjang: are you sure about that? Take $\frac{xy}{x^2+y^2}$; if you compute the limit with $x$ first and then $y$ or with $y$ first and then $x$ it gives $0$ both times, but the limit doesn't exist (take the line $y=x$). –  Javier Badia Nov 1 '12 at 23:56

Hints: For problem $1$, use the fact that $|\sin t|\le |t|$. Then to show that the limit of $\frac{xy}{\sqrt{x^2+y^2}}$ is $0$, switch to polar coordinates.

For problem $3$, it is handy to divide top and bottom by $x^2$, taking care separately of the difficulty when $x=0$.

For problem $2$, write $\tan(x+y)$ in terms of $\cos(x+y)$ and $\sin(x+y)$. Be careful about which side of $\pi/2$ the number $x+y$ is.

For problem $4$, it is useful to adjust the exponent so that it is $\frac{1}{3x^2y}$, or $\frac{1}{x^2y}$.

In $2$ and $4$, you may want to take the logarithm, and calculate the limit of that, though it is not necessary.

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Thanks a lot !!! Great hints ! –  joshua Jul 9 '12 at 6:56

Here I am using Andre’s idea to show that $\frac{xy}{\sqrt{x^2+y^2}}$ has limit $0$ when both $x$ and $y$ tend to $0$. For this we have to show: $$\forall \epsilon>0,∃ \delta>0, \forall (x,y), 0<||(x,y)-(0,0)||<\delta \longrightarrow |\frac{xy}{\sqrt{x^2+y^2}}-0|<\epsilon$$ Firstly, saying that $0<||(x,y)-(0,0)||<\delta$ is equivalent to $\sqrt{x^2+y^2}<\delta$ and therefore both of $|x|, |y|$ are less than $\delta$. If you take $z=max\{|x|,|y|\}$ then you have $z<\delta$ and : $$|\frac{xy}{\sqrt {x^2+y^2}}-0|=\frac{|x||y|}{ \sqrt{x^2+y^2}} ≤ \frac{zz}{ \sqrt{z^2+0}}=z<\delta$$ Now, it is enough to take $\delta$ as $\epsilon$. I hope mine could help you just for the first one. :)

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In this kinds of limits, you can take $r_\alpha(t)=(t,\alpha t)$ and put it into your function to find path wise limit of $f$ when $t$ tends to $0$. After simplifying the original function, if the limit of last expression approaches to zero then probably your original function has limit $0$ at $(0,0)$. Now, you have to use $\epsilon, \delta$ to prove your limit. Note that we see $Lim_{t\rightarrow 0} r_\alpha (t)=(0,0)$. –  B. S. Jul 9 '12 at 7:09
+1 for you! :^) –  amWhy Mar 9 '13 at 2:57

For (4),

$$\lim_{(x,y)\to(0,1)}\left(1+3x^2y\right)^\frac{1}{x^2(1+y)}=\lim_{(x,y)\to(0,1)}\left(\left(1+3x^2y\right)^{\frac1{3x^2y}}\right)^{\frac{3y}{1+y}}\;,$$

where $$\lim_{(x,y)\to(0,1)}\left(1+3x^2y\right)^{\frac1{3x^2y}}$$ should be a familiar limit in the one-variable setting.

A similar trick works in (2) if you write the exponent $\tan(x+y)$ as $\dfrac1{\cos(x+y)}\cdot\sin(x+y)$.

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Thanks a lot !!! –  joshua Jul 9 '12 at 6:56

limit 1. $\displaystyle\lim _ {x \to 0 , y \to 0 } \left| \frac{\sin(xy)}{\sqrt{x^2+y^2}} \right | \leq \displaystyle\lim _ {x \to 0 , y \to 0 } \frac{|xy|}{\sqrt{x^2+y^2}}$ now use inequality $a^2+b^2 \geq 2 |ab|$ $\displaystyle\lim _ {x \to 0 , y \to 0 } \frac{|xy|}{\sqrt{x^2+y^2}} \leq \frac{1}{2} \displaystyle\lim _ {x \to 0 , y \to 0 } \sqrt{x^2+y^2}$ which is going to $0$.

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It'S not that hard :). What you wrote here is mainly only missing a pair of $ enclosing the math. – Jorge Campos Jan 26 '13 at 23:52 @JorgeCampos Can you elaborate? – DaveUM Jan 27 '13 at 8:29 Well, it's replacing, say, log(a^z) by $\log(a^z)\$, etc. You can edit your own answer correcting that. Take a look at tobi.oetiker.ch/lshort/lshort.pdf –  Jorge Campos Jan 27 '13 at 21:46