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With derivatives, why doesn't '$h$' in the denominator not invalidate the whole function as diving by zero?

Why use the derivative and not the symmetric derivative?Second derivative “formula derivation”Solving a limit approaching zero with a complex denominatorWhen to Stop Using L'Hôpital's RuleLimits of norms and deriviative as linear transformationFind all points where $f$ is differetiableWhy $(-1 cdot h) = -1$ when $h$ approaches $0$?Why is $limlimits_xto0+xcot x=1$?Why $lim_uto 0lnleft((1+u)^frac1uright)=lnleft(lim_uto 0(1+u)^frac1uright)$?Is this problem right?

2

1

$begingroup$

The '$h$' in the $fracdfdx$ formula, i.e.,
$$f'(x) = lim_hto0fracf(x + h) - f(x)h$$
stands for an infinitesimal change, correct? Why isn't this '$h$' equal $0$? If $hneq 1$ either, why can it be overlooked as inconsequential in the denominator?

I'm self-learning, stop me if I'm mistaken or if the question has already been asked.

calculus limits derivatives

share|cite|improve this question

edited Mar 17 at 11:47

Rócherz

3,0013821

asked Mar 15 at 16:50

user133876user133876

132

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The term "limit" really matters.
  $endgroup$
  – Randall
  Mar 15 at 16:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can't divide by zero, so we have to think about limits instead.
  $endgroup$
  – Jair Taylor
  Mar 15 at 16:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the given context, $h$ represents a non-zero number such that $x+h$ is in the domain of $f$.
  $endgroup$
  – Michael Hoppe
  Mar 16 at 11:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh wow, thank you Michael! That's really helpful!
  $endgroup$
  – user133876
  Mar 16 at 18:26
  

  
  
  
  

add a comment | 

2

1

$begingroup$

The '$h$' in the $fracdfdx$ formula, i.e.,
$$f'(x) = lim_hto0fracf(x + h) - f(x)h$$
stands for an infinitesimal change, correct? Why isn't this '$h$' equal $0$? If $hneq 1$ either, why can it be overlooked as inconsequential in the denominator?

I'm self-learning, stop me if I'm mistaken or if the question has already been asked.

calculus limits derivatives

share|cite|improve this question

edited Mar 17 at 11:47

Rócherz

3,0013821

asked Mar 15 at 16:50

user133876user133876

132

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The term "limit" really matters.
  $endgroup$
  – Randall
  Mar 15 at 16:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can't divide by zero, so we have to think about limits instead.
  $endgroup$
  – Jair Taylor
  Mar 15 at 16:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In the given context, $h$ represents a non-zero number such that $x+h$ is in the domain of $f$.
  $endgroup$
  – Michael Hoppe
  Mar 16 at 11:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh wow, thank you Michael! That's really helpful!
  $endgroup$
  – user133876
  Mar 16 at 18:26
  

  
  
  
  

add a comment | 

$begingroup$

The '$h$' in the $fracdfdx$ formula, i.e.,
$$f'(x) = lim_hto0fracf(x + h) - f(x)h$$
stands for an infinitesimal change, correct? Why isn't this '$h$' equal $0$? If $hneq 1$ either, why can it be overlooked as inconsequential in the denominator?

I'm self-learning, stop me if I'm mistaken or if the question has already been asked.

calculus limits derivatives

share|cite|improve this question

edited Mar 17 at 11:47

Rócherz

3,0013821

asked Mar 15 at 16:50

user133876user133876

132

$endgroup$

share|cite|improve this question

edited Mar 17 at 11:47

Rócherz

3,0013821

asked Mar 15 at 16:50

user133876user133876

132

share|cite|improve this question

edited Mar 17 at 11:47

Rócherz

3,0013821

asked Mar 15 at 16:50

user133876user133876

132

edited Mar 17 at 11:47

Rócherz

3,0013821

edited Mar 17 at 11:47

Rócherz

3,0013821

asked Mar 15 at 16:50

user133876user133876

132

asked Mar 15 at 16:50

user133876user133876

132

2 Answers
2

active

oldest

votes

4

$begingroup$

You’re correct that we can’t simply divide by $h$ if $h$ is zero. But recall that $frac 0 0$ is an indeterminate form; it does not necessarily diverge to infinity, and in fact when a limit takes on this form it can take on any real value. This is the form of the definition of the derivative: $lim_hto 0fracf(x+h)-f(x)h$ becomes $frac 0 0$ upon plugging in $0$ for $h$, so we cannot simply plug in and evaluate the limit like this. But as we approach $h=0$ via a limit, both the numerator and denominator will shrink towards zero, while the ratio as a whole will approach some other value. This is how we “get around” the issue of the indeterminate form and division by zero.

share|cite|improve this answer

edited Mar 17 at 11:46

Rócherz

3,0013821

answered Mar 15 at 17:38

csch2csch2

6131314

$endgroup$

add a comment | 

0

$begingroup$

Even though dividing by $0$ introduces challenges, it doesn’t necessarily completely “invalidate” what you are doing. Take $1/0$ this can’t be analyzed by itself, but it is still valuable. To “approach” (pun intended) this issue, we use a limit, which upon closer inspection reveals that the limit grows without bound, a much more informative term than DNE, or invalid.

In the case of a derivative, many people describe it as a “paradox” because of exactly your question above, however I think this is wrong, it not a paradox at all, it’s an indeterminate limit as described in the previous answer which demands some sort of “closer look.”

share|cite|improve this answer

answered Mar 15 at 23:32

Roshan Klein-SeetharamanRoshan Klein-Seetharaman

637

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This answer is enlightening as well, thank you!
  $endgroup$
  – user133876
  Mar 16 at 1:14
  

  
  
  
  

add a comment | 

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4

$begingroup$

You’re correct that we can’t simply divide by $h$ if $h$ is zero. But recall that $frac 0 0$ is an indeterminate form; it does not necessarily diverge to infinity, and in fact when a limit takes on this form it can take on any real value. This is the form of the definition of the derivative: $lim_hto 0fracf(x+h)-f(x)h$ becomes $frac 0 0$ upon plugging in $0$ for $h$, so we cannot simply plug in and evaluate the limit like this. But as we approach $h=0$ via a limit, both the numerator and denominator will shrink towards zero, while the ratio as a whole will approach some other value. This is how we “get around” the issue of the indeterminate form and division by zero.

share|cite|improve this answer

edited Mar 17 at 11:46

Rócherz

3,0013821

answered Mar 15 at 17:38

csch2csch2

6131314

$endgroup$

add a comment | 

4

$begingroup$

You’re correct that we can’t simply divide by $h$ if $h$ is zero. But recall that $frac 0 0$ is an indeterminate form; it does not necessarily diverge to infinity, and in fact when a limit takes on this form it can take on any real value. This is the form of the definition of the derivative: $lim_hto 0fracf(x+h)-f(x)h$ becomes $frac 0 0$ upon plugging in $0$ for $h$, so we cannot simply plug in and evaluate the limit like this. But as we approach $h=0$ via a limit, both the numerator and denominator will shrink towards zero, while the ratio as a whole will approach some other value. This is how we “get around” the issue of the indeterminate form and division by zero.

share|cite|improve this answer

edited Mar 17 at 11:46

Rócherz

3,0013821

answered Mar 15 at 17:38

csch2csch2

6131314

$endgroup$

add a comment | 

$begingroup$

You’re correct that we can’t simply divide by $h$ if $h$ is zero. But recall that $frac 0 0$ is an indeterminate form; it does not necessarily diverge to infinity, and in fact when a limit takes on this form it can take on any real value. This is the form of the definition of the derivative: $lim_hto 0fracf(x+h)-f(x)h$ becomes $frac 0 0$ upon plugging in $0$ for $h$, so we cannot simply plug in and evaluate the limit like this. But as we approach $h=0$ via a limit, both the numerator and denominator will shrink towards zero, while the ratio as a whole will approach some other value. This is how we “get around” the issue of the indeterminate form and division by zero.

share|cite|improve this answer

edited Mar 17 at 11:46

Rócherz

3,0013821

answered Mar 15 at 17:38

csch2csch2

6131314

$endgroup$

share|cite|improve this answer

edited Mar 17 at 11:46

Rócherz

3,0013821

answered Mar 15 at 17:38

csch2csch2

6131314

edited Mar 17 at 11:46

Rócherz

3,0013821

edited Mar 17 at 11:46

Rócherz

3,0013821

answered Mar 15 at 17:38

csch2csch2

6131314

answered Mar 15 at 17:38

csch2csch2

6131314

0

$begingroup$

Even though dividing by $0$ introduces challenges, it doesn’t necessarily completely “invalidate” what you are doing. Take $1/0$ this can’t be analyzed by itself, but it is still valuable. To “approach” (pun intended) this issue, we use a limit, which upon closer inspection reveals that the limit grows without bound, a much more informative term than DNE, or invalid.

In the case of a derivative, many people describe it as a “paradox” because of exactly your question above, however I think this is wrong, it not a paradox at all, it’s an indeterminate limit as described in the previous answer which demands some sort of “closer look.”

share|cite|improve this answer

answered Mar 15 at 23:32

Roshan Klein-SeetharamanRoshan Klein-Seetharaman

637

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This answer is enlightening as well, thank you!
  $endgroup$
  – user133876
  Mar 16 at 1:14
  

  
  
  
  

add a comment | 

0

$begingroup$

Even though dividing by $0$ introduces challenges, it doesn’t necessarily completely “invalidate” what you are doing. Take $1/0$ this can’t be analyzed by itself, but it is still valuable. To “approach” (pun intended) this issue, we use a limit, which upon closer inspection reveals that the limit grows without bound, a much more informative term than DNE, or invalid.

In the case of a derivative, many people describe it as a “paradox” because of exactly your question above, however I think this is wrong, it not a paradox at all, it’s an indeterminate limit as described in the previous answer which demands some sort of “closer look.”

share|cite|improve this answer

answered Mar 15 at 23:32

Roshan Klein-SeetharamanRoshan Klein-Seetharaman

637

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This answer is enlightening as well, thank you!
  $endgroup$
  – user133876
  Mar 16 at 1:14
  

  
  
  
  

add a comment | 

$begingroup$

Even though dividing by $0$ introduces challenges, it doesn’t necessarily completely “invalidate” what you are doing. Take $1/0$ this can’t be analyzed by itself, but it is still valuable. To “approach” (pun intended) this issue, we use a limit, which upon closer inspection reveals that the limit grows without bound, a much more informative term than DNE, or invalid.

In the case of a derivative, many people describe it as a “paradox” because of exactly your question above, however I think this is wrong, it not a paradox at all, it’s an indeterminate limit as described in the previous answer which demands some sort of “closer look.”

share|cite|improve this answer

answered Mar 15 at 23:32

Roshan Klein-SeetharamanRoshan Klein-Seetharaman

637

$endgroup$

share|cite|improve this answer

answered Mar 15 at 23:32

Roshan Klein-SeetharamanRoshan Klein-Seetharaman

637

answered Mar 15 at 23:32

Roshan Klein-SeetharamanRoshan Klein-Seetharaman

637

answered Mar 15 at 23:32

Roshan Klein-SeetharamanRoshan Klein-Seetharaman

637