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

would also exist. (Recall that $\dst\lim_{t\rightarrow 0}\tfrac{\sin t}{t} =1$.)