# 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 !

#### Solutions Collecting From Web of "Multivariable Limits"

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.

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 $\dfrac1{\cos(x+y)}\cdot\sin(x+y)$.

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. 🙂

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 $0$.

For #2, If you let t = x+y then the limit becomes lim as t–>pi/2 in t alone.
Now let L = the original limit and take log (ln=log, for me) of both sides using the log rule log (a^z) = a*log(z). Finally (I won’t do the algebra for you) you end up with a limit, in t, which is -1. Keeping in mind that you applied the log, you must now reverse that step. I.E. if the original limit is L you now have log(L) = -1, so L = e^(-1) or 1/e.

This “log trick” is a very useful (and legitimate as long as you avoid log(0)) that brings the exponent sort of “down to earth”. so you can handle things all on one level with much less anxiety! the same trick will work on #4)

(I am sorry I haven’t learned how to make the text look “fancy” by using all the syntactical rules. I keep allowing the mathematics to take preference over the formatting. I would really appreciate some sort of hands on “tutorial” rather than a list of benign rules. I am a “learn by doing” kind of person.)