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2

$begingroup$

Suppose one is asked to explain why the real valued function

$$f(x) = sqrt[3]{x}$$

is not differentiable at $x = 0$ and gives the following argument: For $x neq 0$, one has

$$f'(x) = frac{1}{3sqrt[3]{x^2}}$$

and this is not defined for $x = 0$. Is this a perfectly valid argument? In my opinion, it is not since $f'(0)$ is a limit and it $textit{could}$ be that the limit nevertheless exists. I mean the following: If one says that $f'(0)$ does not exists because $frac{1}{3 cdot 0}$ is not defined, one implicitly assumes that $f'$ is continuous at $x = 0$, or? I think, that the correct answer would be, that

$$lim_{x to 0} f'(x) = infty.$$

(My question arises from the point that a function could have a limit in a point, although it is not defined there. Suppose for example you have $g(x) = frac{x}{x}$. Then $g$ is not defined at $x = 0$ but $lim_{x to 0} g (x) = 1$.)

real-analysis calculus limits

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asked 10 hours ago

JanJan

1,18911 gold badge33 silver badges2121 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is not valid as is. More justification is required to use that argument.
  $endgroup$
  – Don Thousand
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See math.stackexchange.com/questions/97213/… math.stackexchange.com/questions/2356839/…
  $endgroup$
  – thewitness
  9 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Suppose one is asked to explain why the real valued function

$$f(x) = sqrt[3]{x}$$

is not differentiable at $x = 0$ and gives the following argument: For $x neq 0$, one has

$$f'(x) = frac{1}{3sqrt[3]{x^2}}$$

and this is not defined for $x = 0$. Is this a perfectly valid argument? In my opinion, it is not since $f'(0)$ is a limit and it $textit{could}$ be that the limit nevertheless exists. I mean the following: If one says that $f'(0)$ does not exists because $frac{1}{3 cdot 0}$ is not defined, one implicitly assumes that $f'$ is continuous at $x = 0$, or? I think, that the correct answer would be, that

$$lim_{x to 0} f'(x) = infty.$$

(My question arises from the point that a function could have a limit in a point, although it is not defined there. Suppose for example you have $g(x) = frac{x}{x}$. Then $g$ is not defined at $x = 0$ but $lim_{x to 0} g (x) = 1$.)

real-analysis calculus limits

share|cite|improve this question

asked 10 hours ago

JanJan

1,18911 gold badge33 silver badges2121 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is not valid as is. More justification is required to use that argument.
  $endgroup$
  – Don Thousand
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See math.stackexchange.com/questions/97213/… math.stackexchange.com/questions/2356839/…
  $endgroup$
  – thewitness
  9 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Suppose one is asked to explain why the real valued function

$$f(x) = sqrt[3]{x}$$

is not differentiable at $x = 0$ and gives the following argument: For $x neq 0$, one has

$$f'(x) = frac{1}{3sqrt[3]{x^2}}$$

and this is not defined for $x = 0$. Is this a perfectly valid argument? In my opinion, it is not since $f'(0)$ is a limit and it $textit{could}$ be that the limit nevertheless exists. I mean the following: If one says that $f'(0)$ does not exists because $frac{1}{3 cdot 0}$ is not defined, one implicitly assumes that $f'$ is continuous at $x = 0$, or? I think, that the correct answer would be, that

$$lim_{x to 0} f'(x) = infty.$$

(My question arises from the point that a function could have a limit in a point, although it is not defined there. Suppose for example you have $g(x) = frac{x}{x}$. Then $g$ is not defined at $x = 0$ but $lim_{x to 0} g (x) = 1$.)

real-analysis calculus limits

share|cite|improve this question

asked 10 hours ago

JanJan

1,18911 gold badge33 silver badges2121 bronze badges

$endgroup$

share|cite|improve this question

asked 10 hours ago

JanJan

1,18911 gold badge33 silver badges2121 bronze badges

share|cite|improve this question

asked 10 hours ago

JanJan

1,18911 gold badge33 silver badges2121 bronze badges

asked 10 hours ago

JanJan

1,18911 gold badge33 silver badges2121 bronze badges

asked 10 hours ago

JanJan

1,18911 gold badge33 silver badges2121 bronze badges

2 Answers
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4

$begingroup$

That the expression giving $f’(x)$ for $x > 0$ is undefined at $x=0$ is not a valid reason to state that $f$ isn’t differentiable at $0$.

You can think for instance of $f(x)=x^{3/2}sin{frac{1}{x}}$: for $x > 0$, $f’(x)=-x^{-1/2}cos{frac{1}{x}}+3/2x^{1/2}sin{frac{1}{x}}$, so it is undefined at $x=0$.
However, $f’(0)=0$.

On the other hand, if $f’(x) rightarrow infty$ when $x rightarrow 0$, then $f$ isn’t differentiable at $0$. Indeed, for all $x$, $frac{f(x)-f(0)}{x}=f’(c_x)$ for some $0 < c_x < x$, thus $f’(c_x) rightarrow infty$ as $x rightarrow 0$.

share|cite|improve this answer

answered 9 hours ago

MindlackMindlack

6,42855 silver badges1313 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

For checking differentiability of a function,say $f$ at a point $x=a$ one should always use the definition of differentiability i.e. try to check if $$lim_{xto a} frac{f(x)-f(a)}{x-a}$$ exists or not. It may so happen that the limit exists but the derivative function $f'$ is not continuous or defined at that particular point (Consider the classical example of the function: $f(x)=x^2.sin(x)$ when $xneq0$ and $f(x)=0$ when $x=0$).

Important thing to note that if you have a function defined in the interval $(a,b)$ and suppose that $cin(a,b)$ and $f$ is differentiable in $(a,c)$ and in $(c,b)$ and if $$lim_{xto c}f'(x)$$ exists then $f$ is indeed differentiable at $c$ and $$lim_{xto c}f'(x)=f'(c)$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

user587126user587126

7477 bronze badges

$endgroup$

add a comment | 

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4

$begingroup$

That the expression giving $f’(x)$ for $x > 0$ is undefined at $x=0$ is not a valid reason to state that $f$ isn’t differentiable at $0$.

You can think for instance of $f(x)=x^{3/2}sin{frac{1}{x}}$: for $x > 0$, $f’(x)=-x^{-1/2}cos{frac{1}{x}}+3/2x^{1/2}sin{frac{1}{x}}$, so it is undefined at $x=0$.
However, $f’(0)=0$.

On the other hand, if $f’(x) rightarrow infty$ when $x rightarrow 0$, then $f$ isn’t differentiable at $0$. Indeed, for all $x$, $frac{f(x)-f(0)}{x}=f’(c_x)$ for some $0 < c_x < x$, thus $f’(c_x) rightarrow infty$ as $x rightarrow 0$.

share|cite|improve this answer

answered 9 hours ago

MindlackMindlack

6,42855 silver badges1313 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

That the expression giving $f’(x)$ for $x > 0$ is undefined at $x=0$ is not a valid reason to state that $f$ isn’t differentiable at $0$.

You can think for instance of $f(x)=x^{3/2}sin{frac{1}{x}}$: for $x > 0$, $f’(x)=-x^{-1/2}cos{frac{1}{x}}+3/2x^{1/2}sin{frac{1}{x}}$, so it is undefined at $x=0$.
However, $f’(0)=0$.

On the other hand, if $f’(x) rightarrow infty$ when $x rightarrow 0$, then $f$ isn’t differentiable at $0$. Indeed, for all $x$, $frac{f(x)-f(0)}{x}=f’(c_x)$ for some $0 < c_x < x$, thus $f’(c_x) rightarrow infty$ as $x rightarrow 0$.

share|cite|improve this answer

answered 9 hours ago

MindlackMindlack

6,42855 silver badges1313 bronze badges

$endgroup$

add a comment | 

$begingroup$

That the expression giving $f’(x)$ for $x > 0$ is undefined at $x=0$ is not a valid reason to state that $f$ isn’t differentiable at $0$.

You can think for instance of $f(x)=x^{3/2}sin{frac{1}{x}}$: for $x > 0$, $f’(x)=-x^{-1/2}cos{frac{1}{x}}+3/2x^{1/2}sin{frac{1}{x}}$, so it is undefined at $x=0$.
However, $f’(0)=0$.

On the other hand, if $f’(x) rightarrow infty$ when $x rightarrow 0$, then $f$ isn’t differentiable at $0$. Indeed, for all $x$, $frac{f(x)-f(0)}{x}=f’(c_x)$ for some $0 < c_x < x$, thus $f’(c_x) rightarrow infty$ as $x rightarrow 0$.

share|cite|improve this answer

answered 9 hours ago

MindlackMindlack

6,42855 silver badges1313 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

MindlackMindlack

6,42855 silver badges1313 bronze badges

answered 9 hours ago

MindlackMindlack

6,42855 silver badges1313 bronze badges

answered 9 hours ago

MindlackMindlack

6,42855 silver badges1313 bronze badges

1

$begingroup$

For checking differentiability of a function,say $f$ at a point $x=a$ one should always use the definition of differentiability i.e. try to check if $$lim_{xto a} frac{f(x)-f(a)}{x-a}$$ exists or not. It may so happen that the limit exists but the derivative function $f'$ is not continuous or defined at that particular point (Consider the classical example of the function: $f(x)=x^2.sin(x)$ when $xneq0$ and $f(x)=0$ when $x=0$).

Important thing to note that if you have a function defined in the interval $(a,b)$ and suppose that $cin(a,b)$ and $f$ is differentiable in $(a,c)$ and in $(c,b)$ and if $$lim_{xto c}f'(x)$$ exists then $f$ is indeed differentiable at $c$ and $$lim_{xto c}f'(x)=f'(c)$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

user587126user587126

7477 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

For checking differentiability of a function,say $f$ at a point $x=a$ one should always use the definition of differentiability i.e. try to check if $$lim_{xto a} frac{f(x)-f(a)}{x-a}$$ exists or not. It may so happen that the limit exists but the derivative function $f'$ is not continuous or defined at that particular point (Consider the classical example of the function: $f(x)=x^2.sin(x)$ when $xneq0$ and $f(x)=0$ when $x=0$).

Important thing to note that if you have a function defined in the interval $(a,b)$ and suppose that $cin(a,b)$ and $f$ is differentiable in $(a,c)$ and in $(c,b)$ and if $$lim_{xto c}f'(x)$$ exists then $f$ is indeed differentiable at $c$ and $$lim_{xto c}f'(x)=f'(c)$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

user587126user587126

7477 bronze badges

$endgroup$

add a comment | 

$begingroup$

For checking differentiability of a function,say $f$ at a point $x=a$ one should always use the definition of differentiability i.e. try to check if $$lim_{xto a} frac{f(x)-f(a)}{x-a}$$ exists or not. It may so happen that the limit exists but the derivative function $f'$ is not continuous or defined at that particular point (Consider the classical example of the function: $f(x)=x^2.sin(x)$ when $xneq0$ and $f(x)=0$ when $x=0$).

Important thing to note that if you have a function defined in the interval $(a,b)$ and suppose that $cin(a,b)$ and $f$ is differentiable in $(a,c)$ and in $(c,b)$ and if $$lim_{xto c}f'(x)$$ exists then $f$ is indeed differentiable at $c$ and $$lim_{xto c}f'(x)=f'(c)$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

user587126user587126

7477 bronze badges

$endgroup$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

user587126user587126

7477 bronze badges

answered 9 hours ago

user587126user587126

7477 bronze badges

answered 9 hours ago

user587126user587126

7477 bronze badges