Multivariable epsilon delta limit $f(x,y) = 3x + 4y $ as $(x,y) to (1,2)$

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Given that $f(x,y) = 3x+4y$, I need the limit $(x,y) to (2,1)$. I know limit is 10, so
$$|3x+4y -10| = |3(x-2)+4(y-1)| < 3|x-2|+4|y-1|$$

Here we know $|x-2| < delta = sqrt (x-2)^2 + (y-1)^2$ and same for $|y-1|$, so we say inequality is satisfied when

$$7delta lt epsilon$$

or when $delta < epsilon / 7$.

Is this correct proof?

edited Aug 29 at 4:05

asked Aug 29 at 3:59

jeea

45612

add a commentÂ |Â

up vote
0
down vote

favorite

Given that $f(x,y) = 3x+4y$, I need the limit $(x,y) to (2,1)$. I know limit is 10, so
$$|3x+4y -10| = |3(x-2)+4(y-1)| < 3|x-2|+4|y-1|$$

Here we know $|x-2| < delta = sqrt (x-2)^2 + (y-1)^2$ and same for $|y-1|$, so we say inequality is satisfied when

$$7delta lt epsilon$$

or when $delta < epsilon / 7$.

Is this correct proof?

edited Aug 29 at 4:05

asked Aug 29 at 3:59

jeea

45612

add a commentÂ |Â

up vote
0
down vote

favorite

Given that $f(x,y) = 3x+4y$, I need the limit $(x,y) to (2,1)$. I know limit is 10, so
$$|3x+4y -10| = |3(x-2)+4(y-1)| < 3|x-2|+4|y-1|$$

Here we know $|x-2| < delta = sqrt (x-2)^2 + (y-1)^2$ and same for $|y-1|$, so we say inequality is satisfied when

$$7delta lt epsilon$$

or when $delta < epsilon / 7$.

Is this correct proof?

edited Aug 29 at 4:05

asked Aug 29 at 3:59

jeea

45612

Given that $f(x,y) = 3x+4y$, I need the limit $(x,y) to (2,1)$. I know limit is 10, so
$$|3x+4y -10| = |3(x-2)+4(y-1)| < 3|x-2|+4|y-1|$$

Here we know $|x-2| < delta = sqrt (x-2)^2 + (y-1)^2$ and same for $|y-1|$, so we say inequality is satisfied when

$$7delta lt epsilon$$

or when $delta < epsilon / 7$.

Is this correct proof?

edited Aug 29 at 4:05

asked Aug 29 at 3:59

jeea

45612

edited Aug 29 at 4:05

asked Aug 29 at 3:59

jeea

45612

asked Aug 29 at 3:59

jeea

45612

asked Aug 29 at 3:59

jeea

45612

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
2
down vote

accepted

$forall epsilon > 0$, choose $delta = fracepsilon7$, if $sqrt(x-2)^2+(y-1)^2 < delta, $ then
$$|x-2| colorbluele sqrt(x-2)^2+(y-1)^2 < delta$$ and $$|y-1| colorbluele sqrt(x-2)^2+(y-1)^2 < delta$$

which implies that $$|3x+4y -10| = |3(x-2)+4(y-1)| colorbluele 3|x-2|+4|y-1|< 7 delta=epsilon.$$

It is alright to choose a smaller $delta$ too.

answered Aug 29 at 4:12

Siong Thye Goh

81.2k1453103

Thanks a lot sir! I realise the mistake!
â€“Â jeea
Aug 29 at 4:17

add a commentÂ |Â

up vote
2
down vote

You have the correct idea about $$delta < epsilon / 7$$

but I would avoid statements like $$ |x-2| < delta = sqrt (x-2)^2 + (y-1)^2$$
If you rewrite your proof, it will be a very good one.

answered Aug 29 at 4:15

Mohammad Riazi-Kermani

30.8k41852

I realise the error, thanks a lot sir!
â€“Â jeea
Aug 29 at 4:17

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
2
down vote

accepted

$forall epsilon > 0$, choose $delta = fracepsilon7$, if $sqrt(x-2)^2+(y-1)^2 < delta, $ then
$$|x-2| colorbluele sqrt(x-2)^2+(y-1)^2 < delta$$ and $$|y-1| colorbluele sqrt(x-2)^2+(y-1)^2 < delta$$

which implies that $$|3x+4y -10| = |3(x-2)+4(y-1)| colorbluele 3|x-2|+4|y-1|< 7 delta=epsilon.$$

It is alright to choose a smaller $delta$ too.

answered Aug 29 at 4:12

Siong Thye Goh

81.2k1453103

Thanks a lot sir! I realise the mistake!
â€“Â jeea
Aug 29 at 4:17

add a commentÂ |Â

up vote
2
down vote

accepted

$forall epsilon > 0$, choose $delta = fracepsilon7$, if $sqrt(x-2)^2+(y-1)^2 < delta, $ then
$$|x-2| colorbluele sqrt(x-2)^2+(y-1)^2 < delta$$ and $$|y-1| colorbluele sqrt(x-2)^2+(y-1)^2 < delta$$

which implies that $$|3x+4y -10| = |3(x-2)+4(y-1)| colorbluele 3|x-2|+4|y-1|< 7 delta=epsilon.$$

It is alright to choose a smaller $delta$ too.

answered Aug 29 at 4:12

Siong Thye Goh

81.2k1453103

Thanks a lot sir! I realise the mistake!
â€“Â jeea
Aug 29 at 4:17

add a commentÂ |Â

up vote
2
down vote

accepted

$forall epsilon > 0$, choose $delta = fracepsilon7$, if $sqrt(x-2)^2+(y-1)^2 < delta, $ then
$$|x-2| colorbluele sqrt(x-2)^2+(y-1)^2 < delta$$ and $$|y-1| colorbluele sqrt(x-2)^2+(y-1)^2 < delta$$

which implies that $$|3x+4y -10| = |3(x-2)+4(y-1)| colorbluele 3|x-2|+4|y-1|< 7 delta=epsilon.$$

It is alright to choose a smaller $delta$ too.

answered Aug 29 at 4:12

Siong Thye Goh

81.2k1453103

$forall epsilon > 0$, choose $delta = fracepsilon7$, if $sqrt(x-2)^2+(y-1)^2 < delta, $ then
$$|x-2| colorbluele sqrt(x-2)^2+(y-1)^2 < delta$$ and $$|y-1| colorbluele sqrt(x-2)^2+(y-1)^2 < delta$$

which implies that $$|3x+4y -10| = |3(x-2)+4(y-1)| colorbluele 3|x-2|+4|y-1|< 7 delta=epsilon.$$

It is alright to choose a smaller $delta$ too.

answered Aug 29 at 4:12

Siong Thye Goh

81.2k1453103

answered Aug 29 at 4:12

Siong Thye Goh

81.2k1453103

answered Aug 29 at 4:12

Siong Thye Goh

81.2k1453103

answered Aug 29 at 4:12

Siong Thye Goh

81.2k1453103

Thanks a lot sir! I realise the mistake!
â€“Â jeea
Aug 29 at 4:17

add a commentÂ |Â

Thanks a lot sir! I realise the mistake!
â€“Â jeea
Aug 29 at 4:17

Thanks a lot sir! I realise the mistake!
â€“Â jeea
Aug 29 at 4:17

add a commentÂ |Â

up vote
2
down vote

You have the correct idea about $$delta < epsilon / 7$$

but I would avoid statements like $$ |x-2| < delta = sqrt (x-2)^2 + (y-1)^2$$
If you rewrite your proof, it will be a very good one.

answered Aug 29 at 4:15

Mohammad Riazi-Kermani

30.8k41852

I realise the error, thanks a lot sir!
â€“Â jeea
Aug 29 at 4:17

add a commentÂ |Â

up vote
2
down vote

You have the correct idea about $$delta < epsilon / 7$$

but I would avoid statements like $$ |x-2| < delta = sqrt (x-2)^2 + (y-1)^2$$
If you rewrite your proof, it will be a very good one.

answered Aug 29 at 4:15

Mohammad Riazi-Kermani

30.8k41852

I realise the error, thanks a lot sir!
â€“Â jeea
Aug 29 at 4:17

add a commentÂ |Â

up vote
2
down vote

You have the correct idea about $$delta < epsilon / 7$$

but I would avoid statements like $$ |x-2| < delta = sqrt (x-2)^2 + (y-1)^2$$
If you rewrite your proof, it will be a very good one.

answered Aug 29 at 4:15

Mohammad Riazi-Kermani

30.8k41852

You have the correct idea about $$delta < epsilon / 7$$

but I would avoid statements like $$ |x-2| < delta = sqrt (x-2)^2 + (y-1)^2$$
If you rewrite your proof, it will be a very good one.

answered Aug 29 at 4:15

Mohammad Riazi-Kermani

30.8k41852

answered Aug 29 at 4:15

Mohammad Riazi-Kermani

30.8k41852

answered Aug 29 at 4:15

Mohammad Riazi-Kermani

30.8k41852

answered Aug 29 at 4:15

Mohammad Riazi-Kermani

30.8k41852

I realise the error, thanks a lot sir!
â€“Â jeea
Aug 29 at 4:17

add a commentÂ |Â

I realise the error, thanks a lot sir!
â€“Â jeea
Aug 29 at 4:17

I realise the error, thanks a lot sir!
â€“Â jeea
Aug 29 at 4:17

add a commentÂ |Â

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