Why does changing the operator in $lim_hto0$ alter the result of this function?

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

Let $f(x) = |x|$.

Attempting to differentiate $f(x)$ at 0 will fail because the limit does not exist at 0 as the left and right side are unequal.

$$lim_hto0dfracf(0+h)-f(0)h$$

However, my problem is about understanding why we get a different answer on the left side and the right side.

Left side :

$$lim_hto0+fracf(0+h)-f(0)h = lim_hto0+frach-0h = 1$$

Right Side :

$$lim_hto0-fracf(0+h)-f(0)h = lim_hto0-frac-h-0h = -1$$

Why does changing the operator in $lim_hto0$ change the result of these equations? More specifically, why do I have to add $-$ before $h$ on the right side?

limits limits-without-lhopital

share|cite|improve this question

edited Jul 18 at 0:10

asked Jul 16 at 22:21

Cedric Martens

286211

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because they represent different limits? These are one-sided limits. Have you seen these before? See here: tutorial.math.lamar.edu/Classes/CalcI/OneSidedLimits.aspx
  – Alex R.
  Jul 16 at 22:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have your left and right sides backwards. For the left side limit, $h lt 0$
  – Ross Millikan
  Jul 16 at 22:29
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

Let $f(x) = |x|$.

Attempting to differentiate $f(x)$ at 0 will fail because the limit does not exist at 0 as the left and right side are unequal.

$$lim_hto0dfracf(0+h)-f(0)h$$

However, my problem is about understanding why we get a different answer on the left side and the right side.

Left side :

$$lim_hto0+fracf(0+h)-f(0)h = lim_hto0+frach-0h = 1$$

Right Side :

$$lim_hto0-fracf(0+h)-f(0)h = lim_hto0-frac-h-0h = -1$$

Why does changing the operator in $lim_hto0$ change the result of these equations? More specifically, why do I have to add $-$ before $h$ on the right side?

limits limits-without-lhopital

share|cite|improve this question

edited Jul 18 at 0:10

asked Jul 16 at 22:21

Cedric Martens

286211

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because they represent different limits? These are one-sided limits. Have you seen these before? See here: tutorial.math.lamar.edu/Classes/CalcI/OneSidedLimits.aspx
  – Alex R.
  Jul 16 at 22:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have your left and right sides backwards. For the left side limit, $h lt 0$
  – Ross Millikan
  Jul 16 at 22:29
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $f(x) = |x|$.

Attempting to differentiate $f(x)$ at 0 will fail because the limit does not exist at 0 as the left and right side are unequal.

$$lim_hto0dfracf(0+h)-f(0)h$$

However, my problem is about understanding why we get a different answer on the left side and the right side.

Left side :

$$lim_hto0+fracf(0+h)-f(0)h = lim_hto0+frach-0h = 1$$

Right Side :

$$lim_hto0-fracf(0+h)-f(0)h = lim_hto0-frac-h-0h = -1$$

Why does changing the operator in $lim_hto0$ change the result of these equations? More specifically, why do I have to add $-$ before $h$ on the right side?

limits limits-without-lhopital

share|cite|improve this question

edited Jul 18 at 0:10

asked Jul 16 at 22:21

Cedric Martens

286211

Let $f(x) = |x|$.

Attempting to differentiate $f(x)$ at 0 will fail because the limit does not exist at 0 as the left and right side are unequal.

$$lim_hto0dfracf(0+h)-f(0)h$$

However, my problem is about understanding why we get a different answer on the left side and the right side.

Left side :

$$lim_hto0+fracf(0+h)-f(0)h = lim_hto0+frach-0h = 1$$

Right Side :

$$lim_hto0-fracf(0+h)-f(0)h = lim_hto0-frac-h-0h = -1$$

Why does changing the operator in $lim_hto0$ change the result of these equations? More specifically, why do I have to add $-$ before $h$ on the right side?

limits limits-without-lhopital

share|cite|improve this question

edited Jul 18 at 0:10

asked Jul 16 at 22:21

Cedric Martens

286211

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 0:10

edited Jul 18 at 0:10

edited Jul 18 at 0:10

asked Jul 16 at 22:21

Cedric Martens

286211

asked Jul 16 at 22:21

Cedric Martens

286211

asked Jul 16 at 22:21

Cedric Martens

286211

286211

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because they represent different limits? These are one-sided limits. Have you seen these before? See here: tutorial.math.lamar.edu/Classes/CalcI/OneSidedLimits.aspx
  – Alex R.
  Jul 16 at 22:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have your left and right sides backwards. For the left side limit, $h lt 0$
  – Ross Millikan
  Jul 16 at 22:29
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Because they represent different limits? These are one-sided limits. Have you seen these before? See here: tutorial.math.lamar.edu/Classes/CalcI/OneSidedLimits.aspx
  – Alex R.
  Jul 16 at 22:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have your left and right sides backwards. For the left side limit, $h lt 0$
  – Ross Millikan
  Jul 16 at 22:29
  

  
  
  
  

1

1

Because they represent different limits? These are one-sided limits. Have you seen these before? See here: tutorial.math.lamar.edu/Classes/CalcI/OneSidedLimits.aspx
– Alex R.
Jul 16 at 22:22

Because they represent different limits? These are one-sided limits. Have you seen these before? See here: tutorial.math.lamar.edu/Classes/CalcI/OneSidedLimits.aspx
– Alex R.
Jul 16 at 22:22

You have your left and right sides backwards. For the left side limit, $h lt 0$
– Ross Millikan
Jul 16 at 22:29

You have your left and right sides backwards. For the left side limit, $h lt 0$
– Ross Millikan
Jul 16 at 22:29

add a comment | 

1 Answer
1

active

oldest

votes

up vote
3
down vote



accepted

For the right side limit $h gt 0,$ so $|0+h|=h$. For the left side limit, $h lt 0,$ so $|0+h|=-h$. It is just the result of applying the absolute value. I think we are prone to intuitively think variables are positive, but that is not the case for the left side limit.

share|cite|improve this answer

answered Jul 16 at 22:29

Ross Millikan

276k21187352

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853917%2fwhy-does-changing-the-operator-in-lim-h-to0-alter-the-result-of-this-functi%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

For the right side limit $h gt 0,$ so $|0+h|=h$. For the left side limit, $h lt 0,$ so $|0+h|=-h$. It is just the result of applying the absolute value. I think we are prone to intuitively think variables are positive, but that is not the case for the left side limit.

share|cite|improve this answer

answered Jul 16 at 22:29

Ross Millikan

276k21187352

add a comment | 

up vote
3
down vote



accepted

For the right side limit $h gt 0,$ so $|0+h|=h$. For the left side limit, $h lt 0,$ so $|0+h|=-h$. It is just the result of applying the absolute value. I think we are prone to intuitively think variables are positive, but that is not the case for the left side limit.

share|cite|improve this answer

answered Jul 16 at 22:29

Ross Millikan

276k21187352

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

For the right side limit $h gt 0,$ so $|0+h|=h$. For the left side limit, $h lt 0,$ so $|0+h|=-h$. It is just the result of applying the absolute value. I think we are prone to intuitively think variables are positive, but that is not the case for the left side limit.

share|cite|improve this answer

answered Jul 16 at 22:29

Ross Millikan

276k21187352

For the right side limit $h gt 0,$ so $|0+h|=h$. For the left side limit, $h lt 0,$ so $|0+h|=-h$. It is just the result of applying the absolute value. I think we are prone to intuitively think variables are positive, but that is not the case for the left side limit.

share|cite|improve this answer

answered Jul 16 at 22:29

Ross Millikan

276k21187352

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 22:29

Ross Millikan

276k21187352

answered Jul 16 at 22:29

Ross Millikan

276k21187352

answered Jul 16 at 22:29

Ross Millikan

276k21187352

276k21187352

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853917%2fwhy-does-changing-the-operator-in-lim-h-to0-alter-the-result-of-this-functi%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email