2.1 Limits

How does the behaviour of a function far away from $(0,0)$ affect its limit at $(0,0)$?

in general, false.

In general, this is false. Consider $f(x,y)=12-(1-10x)^2-(1-10y)^2$.

We often (somewhat lazily) interpret the limit “$\lim\limits_{(x,y)\to(0,0)}f(x,y)=10$" to mean that, as $(x,y)$ gets closer and closer to the origin, $f(x,y)$ gets closer and closer to 10. This isn't exactly what the definition means, though. The definition tells us that, we can guarantee that $f(x,y)$ be very close to 10 by choosing $(x,y)$ very close to $(0,0)$.

The function $f(x,y)$ can also be very close to 10 for some $(x,y)$'s that are not close to $(0,0)$. Moreover, we don't know how close to $(0,0)$ we have to be in order for $f(x,y)$ to be “very close" to 10.

In this analogy, $f(x,y)$ is the diameter of a particle taken from the position $(x,y)$ in the basin.

(a) the position of the particle in the basin
(b) the position in the basin that the millstone hits
(c) 50 $\mu$m

You can probably solve (a) and (b) by just staring at $f(x,y)$.

(a) along the $x$-axis (b) along the $y$-axis (c) $\lim\limits_{(x,y)\to(0,0)}f(x,y)$ does not exist

Recall $\cos^2\theta-\sin^2\theta=\cos(2\theta)$

(a) $r^2\cos(2\theta)$ (b) $\text{min}=-1,\ \text{max}=1$ (c) $\text{min}=-r^2,\ \text{max}=r^2$ (d) $r<\sqrt\epsilon$

(e) $\lim\limits_{(x,y)\to(0,0)}f(x,y)=0$

Theorem 2.1.6 in the CLP-3 text

$f(a,b)$

By Theorem 2.1.6, $f(x,y)$ is continuous over its domain. The domain of a polynomial is everywhere; in this case, $\mathbb R^2$. So, $f(x,y)$ is continuous at $(a,b)$. By the definition of continuity, $\lim\limits_{(x,y)\to(a,b)}f(x,y)=f(a,b)$.

For parts (b), (c), (d), (e), switch to polar coordinates. For part (f),

(a) $2$ (b) undefined (c) undefined (d) $0$ (e) $0$ (f) $1$

(a) $\dst\lim_{(x,y)\rightarrow(2,-1)}\ \big(xy+x^2\big)=2(-1)+2^2=2$

(b) Switching to polar coordinates,

which does not exist, since, for example,

(c) Switching to polar coordinates,

which does not exist, since, for example,

(d) Switching to polar coordinates,

since $|\cos\theta|\le 1$ for all $\theta$.

(e) Switching to polar coordinates,

Here, we used that

for all $r>0$.

(f) To start, observe that

We may evaluate $\dst\left[\lim_{x\rightarrow 0}\ \frac{\sin x}{x}\right]$ by l'H^opital's rule or by using the definition of the derivative to give

Similarly, we may evaluate $\dst\left[\lim_{y\rightarrow 0}\ \frac{e^y-1}{y}\right]$ by l'H^opital's rule or by using the definition of the derivative to give

So all together

Switch to polar coordinates.

(a) $0$ (b) See the solution.

(a) In polar coordinates, $x=r\cos\theta$, $y=r\sin\theta$, so that

As

we have

As $\dst\lim_{(x,y)\to (0,0)}2r^4=0$, the squeeze theorem yields $\dst\lim_{(x,y)\to(0,0)}\frac{x^8+y^8}{x^4+y^4}=0$.

(b)
In polar coordinates

As $(x,y)\to (0,0)$ the first fraction $\frac{1}{r^2}\to\infty$ but the second factor can take many different values. For example, if we send $(x,y)$ towards the origin along the $y$–axis, i.e. with $\theta=\pm\frac{\pi}{2}$,

but if we send $(x,y)$ towards the origin along the line $y=x$, i.e. with $\theta=\frac{\pi}{4},\frac{5\pi}{4}$,

and if we send $(x,y)$ towards the origin along the line $y=-x$, i.e. with $\theta=-\frac{\pi}{4},\frac{3\pi}{4}$,

So $\frac{xy^5}{x^8+y^{10}}$ does not approach a single value as $(x,y)\to(0,0)$ and the limit does not exist.

(a) Switch to polar coordinates.

(b) What are the limits when (i) $x=0$ and $y\rightarrow 0$ and when (ii) $y=0$ and $x\rightarrow 0$?

(a) $0$

(b) The limit does not exist since the limits (i) $x=0$, $y\rightarrow 0$ and (ii) $y=0$, $x\rightarrow 0$ are different.

(a) In polar coordinates

Since

and $2r\rightarrow 0$ as $r\rightarrow 0$, the limit exists and is $0$.

(b) The limit as we approach $(0,0)$ along the $x$-axis is

On the other hand the limit as we approach $(0,0)$ along the $y$-axis is

These are different, so the limit as $(x,y)\rightarrow 0$ does not exist.

We can gain a more detailed understanding of the behaviour of $\frac{x^2-y^4}{x^2+y^4}$ near the origin by switching to polar coordinates.

Now fix any $\theta$ and let $r\rightarrow 0$ (so that we are approaching the origin along the ray that makes an angle $\theta$ with the positive $x$-axis). If $\cos\theta\ne 0$ (i.e. the ray is not part of the $y$-axis)

But if $\cos\theta= 0$ (i.e. the ray is part of the $y$-axis)

For part (a) switch to polar coordinates. For part (b), switch to polar coordinates centred on $(0,1)$. That is, make the change of variables $x=r\cos\theta$, $y=1+r\sin\theta$.

(a) $2$ (b) The limit does not exist. See the solution.

(a) In polar coordinates $x=r\cos\theta$, $y=r\sin\theta$

As

we have

(b) Since

and, in polar coordinates centred on $(0,1)$, $x=r\cos\theta$, $y=1+r\sin\theta$,

we have that the limit does not exist. For example, if we send $(x,y)$ to $(0,1)$ along the line $y=1$, so that $\theta=0$, we get the limit $0$, while if we send $(x,y)$ to $(0,1)$ along the line $y=x+1$, so that $\theta=\frac{\pi}{4}$, we get the limit $\frac{1}{4}$.

For part (c), does there exist a single number, $L$, with the property that $f(x,y)$ is really close to $L$ for all $(x,y)$ that are really close to $(0,0)$?

(a) $0$ (b) $\frac{1}{2}$ (c) No.

(a) We have

Observe that, if $\sin\theta=0$, then

for all $r\ne 0$. If $\sin\theta\ne 0$,

So the limit $\dst\lim_{r\rightarrow 0^+} \frac{\cos^2\theta\sin\theta}{r^2\cos^4\theta+\sin^2\theta}$ exists (and is finite) for all fixed $\theta$ and

(b) We have

(c) Note that in part (a) we showed that as $(x,y)$ approaches $(0,0)$ along any straight line, $f(x,y)$ approaches the limit zero. In part (b) we have just shown that as $(x,y)$ approaches $(0,0)$ along the parabola $y=x^2$, $f(x,y)$ approaches the limit $\half$, not zero. So $f(x,y)$ takes values very close to $0$, for some $(x,y)$'s that are really near $(0,0)$ and also takes values very close to $\frac{1}{2}$, for other $(x,y)$'s that are really near $(0,0)$. There is no single number, $L$, with the property that $f(x,y)$ is really close to $L$ for all $(x,y)$ that are really close to $(0,0)$. So the limit does not exist.

For part (b), consider the ratio of $\frac{\sin(xy)}{x^2+y^2}$ (from part (b)) and $\frac{xy}{x^2+y^2}$ (from part (a)), and recall that $\dst\lim_{t\rightarrow 0}\tfrac{\sin t}{t}=1$.

For part (d) consider the limits along the positive $x$- and $y$-axes.

(a), (b), (d) Do not exist. See the solutions. (c) $0$

(a) Since, in polar coordinates,

we have that the limit does not exist. For example,

(b) This limit does not exist, since if it were to exist the limit