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How does this work? $| [f(x)−g(x)] − (L−M) | leq | f(x)−L | + | g(x)−M |$

How do epsilon-delta proofs work for limits at negative infinity?Why does this proof of the chain rule not work?In the proof of $lim_{xrightarrow 1}(2x-1) = 1$, why do we choose $delta=epsilon/2$?does epsilon-delta definition of limit presuppose that the function is defined everywhere at (x-delta,x+delta)?Question about limits using delta - epsilon definitionProving $frac{d}{dx}x^2=2x$ by definitionTrouble proving that limit of $f(x) = 1/(2x - 1)$ at $1/2$ doesn't existWhat is the difference between Cauchy's Rule and L'Hôpital's Rule?If $f$ is differntiable at $x_0$, then $f$ is continuous at $x_0$. How does this work?Prove $displaystyle{lim_{x to 2}},x^3 + 1 = 9$ $(delta < 1 text{ or } delta leq 1)?$

1

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My teacher was showing my class a proof for the limit difference rule using the epsilon-delta definition, and nearing the end, he showed us this:
|[f(x)−g(x)] − (L−M)| ≤ |f(x)−L| + |g(x)−M| . I know what the triangular inequality is, and how it works, but on the left, he has a subtraction not a addition. Are there steps he didn't show, or did he just make a mistake?

calculus limits proof-explanation

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edited Mar 9 at 16:39

HK Lee

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asked Mar 9 at 16:35

TimTim

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1

$begingroup$

My teacher was showing my class a proof for the limit difference rule using the epsilon-delta definition, and nearing the end, he showed us this:
|[f(x)−g(x)] − (L−M)| ≤ |f(x)−L| + |g(x)−M| . I know what the triangular inequality is, and how it works, but on the left, he has a subtraction not a addition. Are there steps he didn't show, or did he just make a mistake?

calculus limits proof-explanation

share|cite|improve this question

edited Mar 9 at 16:39

HK Lee

14.1k52360

asked Mar 9 at 16:35

TimTim

62

New contributor

Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

My teacher was showing my class a proof for the limit difference rule using the epsilon-delta definition, and nearing the end, he showed us this:
|[f(x)−g(x)] − (L−M)| ≤ |f(x)−L| + |g(x)−M| . I know what the triangular inequality is, and how it works, but on the left, he has a subtraction not a addition. Are there steps he didn't show, or did he just make a mistake?

calculus limits proof-explanation

share|cite|improve this question

edited Mar 9 at 16:39

HK Lee

14.1k52360

asked Mar 9 at 16:35

TimTim

62

New contributor

Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 9 at 16:39

HK Lee

14.1k52360

asked Mar 9 at 16:35

TimTim

62

New contributor

Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 9 at 16:39

HK Lee

14.1k52360

asked Mar 9 at 16:35

TimTim

62

New contributor

Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 9 at 16:39

HK Lee

14.1k52360

edited Mar 9 at 16:39

HK Lee

14.1k52360

asked Mar 9 at 16:35

TimTim

62

New contributor

Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 9 at 16:35

TimTim

62

2 Answers
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3

$begingroup$

The trick is that

$$|M-g(x)|=|g(x)-M|$$ because $$|M-g(x)|=|(-1)(g(x)-M)|=|-1|cdot|g(x)-M|=|g(x)-M|$$

So that $$|(f(x)-g(x))-(L-M)|=|(f(x)-L)+(M-g(x))|leq |f(x)-L|+|g(x)-M|$$

share|cite|improve this answer

edited Mar 9 at 16:44

answered Mar 9 at 16:38

cansomeonehelpmeoutcansomeonehelpmeout

7,1273935

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1

$begingroup$

The triangle inequality gives
$$begin{align}left|[f(x)-g(x)]-(L-M)right|&=left|[f(x)-L]-[g(x)-M]right|
\&leq left|f(x)-Lright|+left|-[g(x)-M]right|
\&=left|f(x)-Lright|+left|g(x)-Mright|end{align}$$

share|cite|improve this answer

answered Mar 9 at 16:38

DaveDave

9,07911033

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3

$begingroup$

The trick is that

$$|M-g(x)|=|g(x)-M|$$ because $$|M-g(x)|=|(-1)(g(x)-M)|=|-1|cdot|g(x)-M|=|g(x)-M|$$

So that $$|(f(x)-g(x))-(L-M)|=|(f(x)-L)+(M-g(x))|leq |f(x)-L|+|g(x)-M|$$

share|cite|improve this answer

edited Mar 9 at 16:44

answered Mar 9 at 16:38

cansomeonehelpmeoutcansomeonehelpmeout

7,1273935

$endgroup$

add a comment | 

3

$begingroup$

The trick is that

$$|M-g(x)|=|g(x)-M|$$ because $$|M-g(x)|=|(-1)(g(x)-M)|=|-1|cdot|g(x)-M|=|g(x)-M|$$

So that $$|(f(x)-g(x))-(L-M)|=|(f(x)-L)+(M-g(x))|leq |f(x)-L|+|g(x)-M|$$

share|cite|improve this answer

edited Mar 9 at 16:44

answered Mar 9 at 16:38

cansomeonehelpmeoutcansomeonehelpmeout

7,1273935

$endgroup$

add a comment | 

$begingroup$

The trick is that

$$|M-g(x)|=|g(x)-M|$$ because $$|M-g(x)|=|(-1)(g(x)-M)|=|-1|cdot|g(x)-M|=|g(x)-M|$$

So that $$|(f(x)-g(x))-(L-M)|=|(f(x)-L)+(M-g(x))|leq |f(x)-L|+|g(x)-M|$$

share|cite|improve this answer

edited Mar 9 at 16:44

answered Mar 9 at 16:38

cansomeonehelpmeoutcansomeonehelpmeout

7,1273935

$endgroup$

share|cite|improve this answer

edited Mar 9 at 16:44

answered Mar 9 at 16:38

cansomeonehelpmeoutcansomeonehelpmeout

7,1273935

answered Mar 9 at 16:38

cansomeonehelpmeoutcansomeonehelpmeout

7,1273935

answered Mar 9 at 16:38

cansomeonehelpmeoutcansomeonehelpmeout

7,1273935

1

$begingroup$

The triangle inequality gives
$$begin{align}left|[f(x)-g(x)]-(L-M)right|&=left|[f(x)-L]-[g(x)-M]right|
\&leq left|f(x)-Lright|+left|-[g(x)-M]right|
\&=left|f(x)-Lright|+left|g(x)-Mright|end{align}$$

share|cite|improve this answer

answered Mar 9 at 16:38

DaveDave

9,07911033

$endgroup$

add a comment | 

1

$begingroup$

The triangle inequality gives
$$begin{align}left|[f(x)-g(x)]-(L-M)right|&=left|[f(x)-L]-[g(x)-M]right|
\&leq left|f(x)-Lright|+left|-[g(x)-M]right|
\&=left|f(x)-Lright|+left|g(x)-Mright|end{align}$$

share|cite|improve this answer

answered Mar 9 at 16:38

DaveDave

9,07911033

$endgroup$

add a comment | 

$begingroup$

The triangle inequality gives
$$begin{align}left|[f(x)-g(x)]-(L-M)right|&=left|[f(x)-L]-[g(x)-M]right|
\&leq left|f(x)-Lright|+left|-[g(x)-M]right|
\&=left|f(x)-Lright|+left|g(x)-Mright|end{align}$$

share|cite|improve this answer

answered Mar 9 at 16:38

DaveDave

9,07911033

$endgroup$

share|cite|improve this answer

answered Mar 9 at 16:38

DaveDave

9,07911033

answered Mar 9 at 16:38

DaveDave

9,07911033

answered Mar 9 at 16:38

DaveDave

9,07911033