Multivariable Limit Problem

**Alterah** · September 18th 2009, 06:55 PM

I am having some problems with the following problems:
1. $\lim_{\theta\to 0}\frac{sin\theta}{\theta} $

Using l'Hôpital's rule I got that the limit equals.

2. $\lim_{(x,y)\to (0,0)}\frac{sin(x + y)}{x + y} $

I am not really sure how to proceed here. I mean, I could approach along the y axis, then am I able to use l'Hôpital's rule?Also, if I approach along y = x afterward I believe I would get 2 for the limit, whereas along the y-axis would yield 1 provided I could use l'Hôpital's rule. In that case the limit would not exist.

3. $\lim_{(x,y)\to (0,0)}\frac{sin(xy)}{xy} $

On this one I don't really know what to do.

**redsoxfan325** · September 18th 2009, 07:13 PM

Originally Posted by Alterah

I am having some problems with the following problems:
1. $\lim_{\theta\to 0}\frac{sin\theta}{\theta} $

Using l'Hôpital's rule I got that the limit equals.

2. $\lim_{(x,y)\to (0,0)}\frac{sin(x + y)}{x + y} $

I am not really sure how to proceed here. I mean, I could approach along the y axis, then am I able to use l'Hôpital's rule?Also, if I approach along y = x afterward I believe I would get 2 for the limit, whereas along the y-axis would yield 1 provided I could use l'Hôpital's rule. In that case the limit would not exist.

3. $\lim_{(x,y)\to (0,0)}\frac{sin(xy)}{xy} $

On this one I don't really know what to do.

1. L'Hopital's rule allows you to take $\lim_{\theta\to0}\frac{\cos\theta}{1}$ , which is clearly $1$ .

2 & 3. The rule: $\lim_{anything\to0}\frac{\sin(anything)}{anything} =1$ . To cover all possible directions at once (except $x=0$ , which you can check manually), simply let $y=f(x)$ . Since $f(x)$ must pass through the origin, $f(0)=0$ , so both $x+f(x)$ and $x\cdot f(x)$ approach $0$ as $x\to 0$ Therefore both questions fall under this rule and have limits of $1$ .

**Prove It** · September 18th 2009, 07:24 PM

Originally Posted by Alterah

I am having some problems with the following problems:
1. $\lim_{\theta\to 0}\frac{sin\theta}{\theta}$ $ $

Using l'Hôpital's rule I got that the limit equals.

2. $\lim_{(x,y)\to (0,0)}\frac{sin(x + y)}{x + y}$ $ $

I am not really sure how to proceed here. I mean, I could approach along the y axis, then am I able to use l'Hôpital's rule?Also, if I approach along y = x afterward I believe I would get 2 for the limit, whereas along the y-axis would yield 1 provided I could use l'Hôpital's rule. In that case the limit would not exist.

3. $\lim_{(x,y)\to (0,0)}\frac{sin(xy)}{xy}$ $ $

On this one I don't really know what to do.

I wouldn't use L'Hospital's Rule to solve 1.

This is because, to find the derivative of $\sin{x}$ , you need to know the limit $\lim_{x \to 0}\frac{\sin{x}}{x}$ , so you have a circular argument.

You can use the sandwich theorem instead - there is plenty of information you can find on google regarding this limit.

**Alterah** · September 18th 2009, 07:30 PM

Originally Posted by redsoxfan325

1. L'Hopital's rule allows you to take $\lim_{\theta\to0}\frac{\cos\theta}{1}$ , which is clearly $1$ .

2 & 3. The rule: $\lim_{anything\to0}\frac{\sin(anything)}{anything} =1$ . To cover all possible directions at once (except $x=0$ , which you can check manually), simply let $y=f(x)$ . Since $f(x)$ must pass through the origin, $f(0)=0$ , so both $x+f(x)$ and $x\cdot f(x)$ approach $0$ as $x\to 0$ Therefore both questions fall under this rule and have limits of $1$ .

Thanks...I forgot to put that I got the limit equals 1 for question 1. I am trying to follow what you have for part two and three. I see how it can work for part 2. But it seems for part three if we use y = f(x) I feel like we get something along the lines of sin(x*f(x))/(x*f(x)) and we still wind up with zero in the denominator. As far as not using L'Hopital's rule for part 1, I Don't really see why not. It's an indeterminate form of 0/0. So I can apply it to it.

**redsoxfan325** · September 18th 2009, 07:37 PM

Originally Posted by Alterah

Thanks...I forgot to put that I got the limit equals 1 for question 1. I am trying to follow what you have for part two and three. I see how it can work for part 2. But it seems for part three if we use y = f(x) I feel like we get something along the lines of sin(x*f(x))/(x*f(x)) and we still wind up with zero in the denominator.

We do get $\frac{\sin(x\cdot f(x))}{x\cdot f(x)}$ . But that's good, because $x\cdot f(x)\to 0$ as $x\to 0$ , which means we can apply that rule I stated.

Originally Posted by Alterah

As far as not using L'Hopital's rule for part 1, I Don't really see why not. It's an indeterminate form of 0/0. So I can apply it to it.

What Prove It is saying is that in order to take the derivative of $\sin x$ at $0$ using the definition of the derivative, you need to take that limit.

$\lim_{h\to 0}\frac{\sin(h)-\sin(0)}{h-0} = \lim_{h\to0}\frac{\sin h}{h}$

**Alterah** · September 19th 2009, 04:42 AM

Originally Posted by redsoxfan325

We do get $\frac{\sin(x\cdot f(x))}{x\cdot f(x)}$ . But that's good, because $x\cdot f(x)\to 0$ as $x\to 0$ , which means we can apply that rule I stated.

Ok. Thanks for the help, how did the rule come about? I suppose I am wanting to verify for myself that this rule holds. At this point the rule seems like "hand waving." Thanks again.

**redsoxfan325** · September 19th 2009, 09:13 AM

Originally Posted by Alterah

Ok. Thanks for the help, how did the rule come about? I suppose I am wanting to verify for myself that this rule holds. At this point the rule seems like "hand waving." Thanks again.

L'Hopital's Rule shows that it works, though you probably wouldn't want to use it in a proof, for the reasons discussed above.

$\lim_{anything\to0}\frac{\sin(anything)}{anything} =\lim_{anything\to0}\frac{\cos(anything)\cdot\frac {d}{dx}[anything]}{\frac{d}{dx}[anything]}$ $=\lim_{anything\to0}\frac{\cos(anything)}{1}$ , which is clearly $1$ .

More officially, you'd probably use something like $u(x)$ instead of "anything", i.e.

$\lim_{u(x)\to0}\frac{\sin(u(x))}{u(x)}=1$

but I used "anything" to try to make it easier to understand.

**Alterah** · September 19th 2009, 09:50 AM

Originally Posted by redsoxfan325

L'Hopital's Rule shows that it works, though you probably wouldn't want to use it in a proof, for the reasons discussed above.

$\lim_{anything\to0}\frac{\sin(anything)}{anything} =\lim_{anything\to0}\frac{\cos(anything)\cdot\frac {d}{dx}[anything]}{\frac{d}{dx}[anything]}$ $=\lim_{anything\to0}\frac{\cos(anything)}{1}$ , which is clearly $1$ .

More officially, you'd probably use something like $u(x)$ instead of "anything", i.e.

$\lim_{u(x)\to0}\frac{\sin(u(x))}{u(x)}=1$

but I used "anything" to try to make it easier to understand.

I see...I suppose it's a bit of a challenge to see with multivariable functions because we actually describe partial derivatives. It makes absolute sense with f(x). Alternatively using the rule I'd think you can get the following:

$\lim_{u(x)\to0}\frac{\cos(u(x))}{u(x)}=0$

correct?

**mr fantastic** · September 19th 2009, 05:46 PM

Originally Posted by Alterah

I see...I suppose it's a bit of a challenge to see with multivariable functions because we actually describe partial derivatives. It makes absolute sense with f(x). Alternatively using the rule I'd think you can get the following:

$\lim_{u(x)\to0}\frac{\cos(u(x))}{u(x)}=0$

correct?

NO!

l'Hopital's can only be used when your limit has the indeterminant form of either $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . I doubt $\lim_{u(x)\to 0}\frac{\cos(u(x))}{u(x)}$ will have either of these two forms, regardless of what u(x) is ....

**Alterah** · September 19th 2009, 06:37 PM

Originally Posted by mr fantastic

NO!

l'Hopital's can only be used when your limit has the indeterminant form of either $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . I doubt $\lim_{u(x)\to 0}\frac{\cos(u(x))}{u(x)}$ will have either of these two forms, regardless of what u(x) is ....

True...I completely forgot about that when I said that about cosine.

**bkarpuz** · September 20th 2009, 01:09 AM

Originally Posted by Alterah

I am having some problems with the following problems:
1. $\lim_{\theta\to 0}\frac{sin\theta}{\theta} $

Using l'Hôpital's rule I got that the limit equals.

2. $\lim_{(x,y)\to (0,0)}\frac{sin(x + y)}{x + y} $

I am not really sure how to proceed here. I mean, I could approach along the y axis, then am I able to use l'Hôpital's rule?Also, if I approach along y = x afterward I believe I would get 2 for the limit, whereas along the y-axis would yield 1 provided I could use l'Hôpital's rule. In that case the limit would not exist.

3. $\lim_{(x,y)\to (0,0)}\frac{sin(xy)}{xy} $

On this one I don't really know what to do.

First I will talk about 2 and 3, which follows from 1.
Clearly, for 2, if $(x,y)\to(0,0)$ , then $x+y\to0$ , and similarly for 3.
Now return to 1.
Let $C$ be the unit ball (hence its area is $\pi$ ).
Draw an equilateral triangle in $C$ such that its corners touch $C$ , denote by $A_{3}$ the area of the triangle.
Then $A_{3}=3\frac{1}{2}\sin\bigg(\frac{2\pi}{3}\bigg)$ .

Now draw a square in $C$ , again let it corners touch $C$ , denote by $A_{4}$ the area of the square.
Then $A_{4}=4\frac{1}{2}\sin\bigg(\frac{2\pi}{4}\bigg)$ .

Similarly, draw a polygon (with $n$ edges of the same lenght), and by $A_{n}$ denote the area of the polygon.
As the number of the corners (edges) increase, the area of the polygon tends to the area of the unit ball (exactly $\pi$ ), i.e.,
$\pi=\lim_{n\to\infty}A_{n}=\lim_{n\to\infty}\frac{ n}{2}\sin\bigg(\dfrac{2\pi}{n}\bigg)$
or equivalently
$\lim_{n\to\infty}\frac{n}{2\pi}\sin\bigg(\dfrac{2\ pi}{n}\bigg)=1\qquad(*)$ .
Now let $u:=2\pi/n$ , and as $n\to\infty$ , we have $u\to0$ , therefore (*) takes the following form
$\lim_{u\to0}\frac{\sin(u)}{u}=1$ .

I hope this helps. :]

**Prove It** · September 20th 2009, 01:37 AM

Alternatively, look at the unit circle.

It should be pretty clear from the diagram that the area of the sector is squeezed between the area of the two triangles made by $\sin{\theta}$ and $\tan{\theta}$ .

So we have

$\frac{1}{2}\sin{\theta}\cos{\theta} \leq \frac{1}{2}\theta \leq \frac{1}{2}\tan{\theta}$

$\sin{\theta}\cos{\theta} \leq \theta \leq \frac{\sin{\theta}}{\cos{\theta}}$

$\frac{\cos{\theta}}{\sin{\theta}} \leq \frac{1}{\theta} \leq \frac{1}{\sin{\theta}\cos{\theta}}$

$\cos{\theta} \leq \frac{\sin{\theta}}{\theta} \leq \frac{1}{\cos{\theta}}$ .

Now apply the sandwich theorem by letting $\theta \to 0$ .

**Alterah** · September 20th 2009, 09:44 AM

Yes, both of those helped immensely. Thanks for all your help!

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