Limits help

i don't think my answer is right, but can some1 just check it?

Also if my answer is wrong, how show I find the limit as (x,y)----(0,0)

For e^-[(x/y)^2] I found out that it is bounded between 0 and 1

Btw all the powers 2 in image

(edited 9 years ago)

image.jpg178.9KB

Reply 1

ztibor

Original post by Vorsah

The limit of

e^{-(\frac{x}{y})^2}

is not exist at (0,0)

Using x-> 0 and y=mx approaching

\lim_{x \rightarrow 0} e^{-(\frac{x}{mx})^2}=e^{-\frac{1}{m^2}}

which is depend on value of m, so the limit is not exist

with another method:

using the

x_n=r\cdot \cos \phi_n

and

y_n=r\cdot \sin \phi_n

sequences as r->0 and

n \rightarrow \infty

\lim_{r \rightarrow 0} e^{-(\frac{r\cdot \cos \phi_n}{r\cdot \sin \phi_n})^2}=e^{-\cot^2 \phi_n}

which value is depend on phi so this limit is not exist.

Reply 2

Vorsah

So is what I done correct?

Reply 3

Vorsah

Original post by ztibor

The limit of

e^{-(\frac{x}{y})^2}

is not exist at (0,0)

Using x-> 0 and y=mx approaching

\lim_{x \rightarrow 0} e^{-(\frac{x}{mx})^2}=e^{-\frac{1}{m^2}}

which is depend on value of m, so the limit is not exist

with another method:

using the

x_n=r\cdot \cos \phi_n

and

y_n=r\cdot \sin \phi_n

sequences as r->0 and

n \rightarrow \infty

\lim_{r \rightarrow 0} e^{-(\frac{r\cdot \cos \phi_n}{r\cdot \sin \phi_n})^2}=e^{-\cot^2 \phi_n}

which value is depend on phi so this limit is not exist.

So is what I've done correct?

Reply 4

ztibor

Original post by Vorsah

So is what I've done correct?

THe limit of the other f(x,y) funtion is correct (0)

but the limit of the product of these two function is not exist

THte product rule for the limit is valid for that both lim g(x,y) and limf(x,y) is exist
and finite.

Reply 5

Vorsah

Original post by ztibor

But I checked the answer, and it says that the limit exist and is 0, since the limit of the other function is 0?

Reply 6

ztibor

Original post by Vorsah

But I checked the answer, and it says that the limit exist and is 0, since the limit of the other function is 0?

Can you post the original question?

Reply 7

Hasufel

This question is interesting -

this may be exceptionally retarded on my part (and please tell me if my logic is rubbish - as this is the way we learn, right? - haven`t dealt too much in multivariable limits!)

but, for the exponential part, i believe the limit to be

Unparseable latex formula:

\diaplaystyle \frac{1}{e}

my reasoning is, treat the exponential part as a separate function, and call it w:

\displaystyle w=e^{-(x/y)^{2}}

taking natural logs gives:

\displaystyle ln(w)=-(x/y)^{2}

and use

lim_{(y->x, x->0)} -(x/y)^{2}= -1

so that

\displaystyle w -> e^{-1}

(edited 9 years ago)

Reply 8

Vorsah

Original post by ztibor

Can you post the original question?

The original Q is, discuss the limit as (x,y)---->(0,0). Provide argument to show whether the limit exists or if the limit does not exist

Reply 9

ztibor

Original post by Hasufel

Unparseable latex formula:

\diaplaystyle \frac{1}{e}

my reasoning is, treat the exponential part as a separate function, and call it w:

\displaystyle w=e^{-(x/y)^{2}}

taking natural logs gives:

\displaystyle ln(w)=-(x/y)^{2}

and use

lim_{(y->x, x->0)} -(x/y)^{2}= -1

so that

\displaystyle w -> e^{-1}

Well, You approache the (0,0) so that x->0 and y=x
that is we approaches (0,0) along y=x line.
- the logarithmic limit is -1;
But:
- Approaching to (0,0) along y=2x as x->0 then this limit will be -1/4
and
- Approaching to (0,0) along y=mx as x->0 then this llimit will be -1/m^2

that is the limit is not unique depends on the value of m
so does not exist

Reply 10

Vorsah

Original post by ztibor

Hey, I still don't understand what the correct answer is you're saying the limit doesn't exist whereas the mark scheme says the limit is 0?

Reply 11

ztibor

Original post by Vorsah

Hey, I still don't understand what the correct answer is you're saying the limit doesn't exist whereas the mark scheme says the limit is 0?

the original question was

\lim_{(x,y)\rightarrow (0,0)} h(x,y)

where

h(x,y)=e^{-(\frac{x}{y})^2}\cdot ln \frac{1-x^2-y^2}{1+x^2+y^2}=g(x,y)\cdot f(x,y)

(if I saw correctly)

You used the multiplication law for this limit so, that

g(x,y)=e^{-(\frac{x}{y})^2}

and

f(x,y)=ln \frac{1-x^2-y^2}{1+x^2+y^2}

The multiplication law says

If the limits

\lim_{x\rightarrow a} f(x)

and

\lim_{x\rightarrow a} g(x)

both exists then

\lim_{x\rightarrow a} f(x)\cdot g(x)=\lim_{x\rightarrow a} f(x) \cdot \lim_{x\rightarrow a} g(x)

In this question:
1.

\lim_{(x,y)\rightarrow (0,0)} f(x,y) =0

\lim_{(x,y)\rightarrow (0,0)} g(x,y)

does not exist
so
3.

\lim_{(x,y)\rightarrow (0,0)} h(x,y)

does not exist (you can not use the multiplication law)

For the mark scheme:

If the mark scheme says the first limit is 0 then that is correct
If the mark schene says any from the other two limits is 0 then that is incorrect

(edited 9 years ago)

Reply 12

Smaug123

Original post by Hasufel

Unparseable latex formula:

\diaplaystyle \frac{1}{e}

my reasoning is, treat the exponential part as a separate function, and call it w:

\displaystyle w=e^{-(x/y)^{2}}

taking natural logs gives:

\displaystyle ln(w)=-(x/y)^{2}

and use

lim_{(y->x, x->0)} -(x/y)^{2}= -1

so that

\displaystyle w -> e^{-1}

\displaystyle \lim_{y \to 0, x \to 0} -\left(\frac{x}{y} \right)^2

doesn't exist - approach along the line x=0 to obtain the limit 0, but approach along the line x=y to obtain the limit -1. (As ztibor says.)

For the limit to exist, we require that the limit be the same when approached not just along any line, but along any curve. In this case, however, two different lines suffice to show the result.

Note that if the function were continuous, then the limit would exist, because that remains a property of continuous functions when we move to more dimensions.

Reply 13

atsruser

Original post by ztibor

\lim_{(x,y)\rightarrow (0,0)} h(x,y)

This looks like dubious reasoning to me.

For example, consider

\lim_{x \rightarrow \infty} e^{-x}\sin x = 0

(by the squeeze/sandwich theorem.)

In this case,

\lim_{x \rightarrow \infty} \sin x

does not exist, but the limit of the product exists just fine. You can't, however, find it by applying the laws of limits. (Maybe you can find an analogue of the law where one limit is 0, and lim sup/lim inf of the other function both exist)

Reply 14

ztibor

Original post by atsruser

This looks like dubious reasoning to me.

For example, consider

\lim_{x \rightarrow \infty} e^{-x}\sin x = 0

(by the squeeze/sandwich theorem.)

In this case the vakues of the

\sin x

are between the finite -1 and 1.
so that always will be a finite value

Do you think that the limit value of

e^{-(\frac{x}{y})^2}

surface always will
be a finite value as (x,y) -> (0,0) ?
What do you think about the

\lim_{x \rightarrow \infty} e^{-x} \frac{1}{\sin x}

limit ?

Reply 15

Vorsah

Original post by ztibor

the original question was

\lim_{(x,y)\rightarrow (0,0)} h(x,y)

where

h(x,y)=e^{-(\frac{x}{y})^2}\cdot ln \frac{1-x^2-y^2}{1+x^2+y^2}=g(x,y)\cdot f(x,y)

(if I saw correctly)

You used the multiplication law for this limit so, that

g(x,y)=e^{-(\frac{x}{y})^2}

and

f(x,y)=ln \frac{1-x^2-y^2}{1+x^2+y^2}

The multiplication law says

If the limits

\lim_{x\rightarrow a} f(x)

and

\lim_{x\rightarrow a} g(x)

both exists then

\lim_{x\rightarrow a} f(x)\cdot g(x)=\lim_{x\rightarrow a} f(x) \cdot \lim_{x\rightarrow a} g(x)

In this question:
1.

\lim_{(x,y)\rightarrow (0,0)} f(x,y) =0

\lim_{(x,y)\rightarrow (0,0)} g(x,y)

does not exist
so
3.

\lim_{(x,y)\rightarrow (0,0)} h(x,y)

The mark scheme says that the exponential function is bounded between 0 and 1. The limit of h(x,y) is 0 as the limit of the logarithmic function is 0 as (x,y)-->(0,0)

(edited 9 years ago)

Reply 16

DFranklin

Original post by ztibor

In this case the vakues of the

\sin x

are between the finite -1 and 1.
so that always will be a finite value

Do you think that the limit value of

e^{-(\frac{x}{y})^2}

surface always will
be a finite value as (x,y) -> (0,0) ?Er, yes? It's the exponential of a non-positive number, so it must lie between 0 and 1.

Edit: there's a question mark about what happens when y = 0 (with x =/= 0), but I'm assuming the actual question either dealt with this or we weren't supposed to worry about it.