Find $lambda$ such that $f$ it is differentiable in zero and has a continuous derivative in zero












2












$begingroup$


I am trying to solve this task




Find $lambda>0$ such that $f=begin{cases}0& x=0\ |x|^{lambda}cdot sinfrac{1}{x} & xneq 0 end{cases}$



a) is differentiable in zero



b) has a continuous derivative in zero.




My try



a)
$$lim_{h rightarrow 0^+}frac{f(h)-f(0)}{h} = lim_{h rightarrow 0^+}frac{|h|^{lambda}}{h} cdot sinfrac{1}{h} = 0$$



$$lim_{h rightarrow 0^-}frac{f(h)-f(0)}{h} = lim_{h rightarrow 0^+}frac{|h|^{lambda}}{h} cdot sinfrac{1}{h} = 0$$
But there I have some doubts. It seems like $lambda$ does not matter there....


Moreover, can somebody give me hints to b)?
I think that I have do the same thing but this is unbelievable






real-analysis calculus limits derivatives continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you think the first limit exists if $lambda = 1/2?$
    $endgroup$
    – zhw.
    Jan 23 at 0:09






  • 1




    $begingroup$
    No, it cannot exists - Does mean that $lambda - 1 > 0 $ ?
    $endgroup$
    – VirtualUser
    Jan 23 at 0:29










  • $begingroup$
    and in the second case we have the same situation?
    $endgroup$
    – VirtualUser
    Jan 23 at 0:29
















2












$begingroup$


I am trying to solve this task




Find $lambda>0$ such that $f=begin{cases}0& x=0\ |x|^{lambda}cdot sinfrac{1}{x} & xneq 0 end{cases}$



a) is differentiable in zero



b) has a continuous derivative in zero.




My try



a)
$$lim_{h rightarrow 0^+}frac{f(h)-f(0)}{h} = lim_{h rightarrow 0^+}frac{|h|^{lambda}}{h} cdot sinfrac{1}{h} = 0$$



$$lim_{h rightarrow 0^-}frac{f(h)-f(0)}{h} = lim_{h rightarrow 0^+}frac{|h|^{lambda}}{h} cdot sinfrac{1}{h} = 0$$
But there I have some doubts. It seems like $lambda$ does not matter there....


Moreover, can somebody give me hints to b)?
I think that I have do the same thing but this is unbelievable






real-analysis calculus limits derivatives continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you think the first limit exists if $lambda = 1/2?$
    $endgroup$
    – zhw.
    Jan 23 at 0:09






  • 1




    $begingroup$
    No, it cannot exists - Does mean that $lambda - 1 > 0 $ ?
    $endgroup$
    – VirtualUser
    Jan 23 at 0:29










  • $begingroup$
    and in the second case we have the same situation?
    $endgroup$
    – VirtualUser
    Jan 23 at 0:29














2












2








2





$begingroup$


I am trying to solve this task




Find $lambda>0$ such that $f=begin{cases}0& x=0\ |x|^{lambda}cdot sinfrac{1}{x} & xneq 0 end{cases}$



a) is differentiable in zero



b) has a continuous derivative in zero.




My try



a)
$$lim_{h rightarrow 0^+}frac{f(h)-f(0)}{h} = lim_{h rightarrow 0^+}frac{|h|^{lambda}}{h} cdot sinfrac{1}{h} = 0$$



$$lim_{h rightarrow 0^-}frac{f(h)-f(0)}{h} = lim_{h rightarrow 0^+}frac{|h|^{lambda}}{h} cdot sinfrac{1}{h} = 0$$
But there I have some doubts. It seems like $lambda$ does not matter there....


Moreover, can somebody give me hints to b)?
I think that I have do the same thing but this is unbelievable






real-analysis calculus limits derivatives continuity







share|cite|improve this question











$endgroup$




I am trying to solve this task




Find $lambda>0$ such that $f=begin{cases}0& x=0\ |x|^{lambda}cdot sinfrac{1}{x} & xneq 0 end{cases}$



a) is differentiable in zero



b) has a continuous derivative in zero.




My try



a)
$$lim_{h rightarrow 0^+}frac{f(h)-f(0)}{h} = lim_{h rightarrow 0^+}frac{|h|^{lambda}}{h} cdot sinfrac{1}{h} = 0$$



$$lim_{h rightarrow 0^-}frac{f(h)-f(0)}{h} = lim_{h rightarrow 0^+}frac{|h|^{lambda}}{h} cdot sinfrac{1}{h} = 0$$
But there I have some doubts. It seems like $lambda$ does not matter there....


Moreover, can somebody give me hints to b)?
I think that I have do the same thing but this is unbelievable






real-analysis calculus limits derivatives continuity




real-analysis calculus limits derivatives continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 1:45









Thomas Shelby

3,7192525




3,7192525










asked Jan 23 at 0:02









VirtualUserVirtualUser

899114




899114












  • $begingroup$
    Do you think the first limit exists if $lambda = 1/2?$
    $endgroup$
    – zhw.
    Jan 23 at 0:09






  • 1




    $begingroup$
    No, it cannot exists - Does mean that $lambda - 1 > 0 $ ?
    $endgroup$
    – VirtualUser
    Jan 23 at 0:29










  • $begingroup$
    and in the second case we have the same situation?
    $endgroup$
    – VirtualUser
    Jan 23 at 0:29


















  • $begingroup$
    Do you think the first limit exists if $lambda = 1/2?$
    $endgroup$
    – zhw.
    Jan 23 at 0:09






  • 1




    $begingroup$
    No, it cannot exists - Does mean that $lambda - 1 > 0 $ ?
    $endgroup$
    – VirtualUser
    Jan 23 at 0:29










  • $begingroup$
    and in the second case we have the same situation?
    $endgroup$
    – VirtualUser
    Jan 23 at 0:29
















$begingroup$
Do you think the first limit exists if $lambda = 1/2?$
$endgroup$
– zhw.
Jan 23 at 0:09




$begingroup$
Do you think the first limit exists if $lambda = 1/2?$
$endgroup$
– zhw.
Jan 23 at 0:09




1




1




$begingroup$
No, it cannot exists - Does mean that $lambda - 1 > 0 $ ?
$endgroup$
– VirtualUser
Jan 23 at 0:29




$begingroup$
No, it cannot exists - Does mean that $lambda - 1 > 0 $ ?
$endgroup$
– VirtualUser
Jan 23 at 0:29












$begingroup$
and in the second case we have the same situation?
$endgroup$
– VirtualUser
Jan 23 at 0:29




$begingroup$
and in the second case we have the same situation?
$endgroup$
– VirtualUser
Jan 23 at 0:29










1 Answer
1






active

oldest

votes


















1












$begingroup$

For the existence part, to make sure the limit $$ f'(0)=lim_{hto 0} frac{|h|^{lambda}}{h}sinfrac{1}{h} $$ exists, we have $ lambda>1 $. And the limit is $ 0 $.



For the continuous derivative part, we have to let the following limit equal to $ 0 $, i.e. :
$$ lim_{hto 0} f'(h)-f'(0)=lim_{hto 0} f'(h)-0=lim_{hto 0} f'(h)=0 .$$
Thus, $$ begin{align} lim_{hto 0+}f'(h)&= lim_{hto 0+}left[(lambda-1)h^{lambda-2}sinfrac{1}{h}-h^{lambda-3}cosfrac{1}{h}right]=0 end{align} $$
and
$$ begin{align} lim_{hto 0-}f'(h)&= -lim_{hto 0-}left[(-lambda+1)(-h)^{lambda-2}sinfrac{1}{h}-(-h)^{lambda-3}cosfrac{1}{h}right]=0 end{align} .$$
Therefore, $ lambda> 3 $.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $ lambda> 3 $ - are you sure about that part? I think that there should be $ lambda> 2 $
    $endgroup$
    – VirtualUser
    Jan 27 at 22:40










  • $begingroup$
    @VirtualUser Yes, since $ cosfrac{1}{h} $ does not exist.
    $endgroup$
    – user549397
    Jan 28 at 0:35











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1 Answer
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1 Answer
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active

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active

oldest

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active

oldest

votes









1












$begingroup$

For the existence part, to make sure the limit $$ f'(0)=lim_{hto 0} frac{|h|^{lambda}}{h}sinfrac{1}{h} $$ exists, we have $ lambda>1 $. And the limit is $ 0 $.



For the continuous derivative part, we have to let the following limit equal to $ 0 $, i.e. :
$$ lim_{hto 0} f'(h)-f'(0)=lim_{hto 0} f'(h)-0=lim_{hto 0} f'(h)=0 .$$
Thus, $$ begin{align} lim_{hto 0+}f'(h)&= lim_{hto 0+}left[(lambda-1)h^{lambda-2}sinfrac{1}{h}-h^{lambda-3}cosfrac{1}{h}right]=0 end{align} $$
and
$$ begin{align} lim_{hto 0-}f'(h)&= -lim_{hto 0-}left[(-lambda+1)(-h)^{lambda-2}sinfrac{1}{h}-(-h)^{lambda-3}cosfrac{1}{h}right]=0 end{align} .$$
Therefore, $ lambda> 3 $.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $ lambda> 3 $ - are you sure about that part? I think that there should be $ lambda> 2 $
    $endgroup$
    – VirtualUser
    Jan 27 at 22:40










  • $begingroup$
    @VirtualUser Yes, since $ cosfrac{1}{h} $ does not exist.
    $endgroup$
    – user549397
    Jan 28 at 0:35
















1












$begingroup$

For the existence part, to make sure the limit $$ f'(0)=lim_{hto 0} frac{|h|^{lambda}}{h}sinfrac{1}{h} $$ exists, we have $ lambda>1 $. And the limit is $ 0 $.



For the continuous derivative part, we have to let the following limit equal to $ 0 $, i.e. :
$$ lim_{hto 0} f'(h)-f'(0)=lim_{hto 0} f'(h)-0=lim_{hto 0} f'(h)=0 .$$
Thus, $$ begin{align} lim_{hto 0+}f'(h)&= lim_{hto 0+}left[(lambda-1)h^{lambda-2}sinfrac{1}{h}-h^{lambda-3}cosfrac{1}{h}right]=0 end{align} $$
and
$$ begin{align} lim_{hto 0-}f'(h)&= -lim_{hto 0-}left[(-lambda+1)(-h)^{lambda-2}sinfrac{1}{h}-(-h)^{lambda-3}cosfrac{1}{h}right]=0 end{align} .$$
Therefore, $ lambda> 3 $.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $ lambda> 3 $ - are you sure about that part? I think that there should be $ lambda> 2 $
    $endgroup$
    – VirtualUser
    Jan 27 at 22:40










  • $begingroup$
    @VirtualUser Yes, since $ cosfrac{1}{h} $ does not exist.
    $endgroup$
    – user549397
    Jan 28 at 0:35














1












1








1





$begingroup$

For the existence part, to make sure the limit $$ f'(0)=lim_{hto 0} frac{|h|^{lambda}}{h}sinfrac{1}{h} $$ exists, we have $ lambda>1 $. And the limit is $ 0 $.



For the continuous derivative part, we have to let the following limit equal to $ 0 $, i.e. :
$$ lim_{hto 0} f'(h)-f'(0)=lim_{hto 0} f'(h)-0=lim_{hto 0} f'(h)=0 .$$
Thus, $$ begin{align} lim_{hto 0+}f'(h)&= lim_{hto 0+}left[(lambda-1)h^{lambda-2}sinfrac{1}{h}-h^{lambda-3}cosfrac{1}{h}right]=0 end{align} $$
and
$$ begin{align} lim_{hto 0-}f'(h)&= -lim_{hto 0-}left[(-lambda+1)(-h)^{lambda-2}sinfrac{1}{h}-(-h)^{lambda-3}cosfrac{1}{h}right]=0 end{align} .$$
Therefore, $ lambda> 3 $.






share|cite|improve this answer









$endgroup$



For the existence part, to make sure the limit $$ f'(0)=lim_{hto 0} frac{|h|^{lambda}}{h}sinfrac{1}{h} $$ exists, we have $ lambda>1 $. And the limit is $ 0 $.



For the continuous derivative part, we have to let the following limit equal to $ 0 $, i.e. :
$$ lim_{hto 0} f'(h)-f'(0)=lim_{hto 0} f'(h)-0=lim_{hto 0} f'(h)=0 .$$
Thus, $$ begin{align} lim_{hto 0+}f'(h)&= lim_{hto 0+}left[(lambda-1)h^{lambda-2}sinfrac{1}{h}-h^{lambda-3}cosfrac{1}{h}right]=0 end{align} $$
and
$$ begin{align} lim_{hto 0-}f'(h)&= -lim_{hto 0-}left[(-lambda+1)(-h)^{lambda-2}sinfrac{1}{h}-(-h)^{lambda-3}cosfrac{1}{h}right]=0 end{align} .$$
Therefore, $ lambda> 3 $.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 23 at 1:36









user549397user549397

1,5081418




1,5081418












  • $begingroup$
    $ lambda> 3 $ - are you sure about that part? I think that there should be $ lambda> 2 $
    $endgroup$
    – VirtualUser
    Jan 27 at 22:40










  • $begingroup$
    @VirtualUser Yes, since $ cosfrac{1}{h} $ does not exist.
    $endgroup$
    – user549397
    Jan 28 at 0:35


















  • $begingroup$
    $ lambda> 3 $ - are you sure about that part? I think that there should be $ lambda> 2 $
    $endgroup$
    – VirtualUser
    Jan 27 at 22:40










  • $begingroup$
    @VirtualUser Yes, since $ cosfrac{1}{h} $ does not exist.
    $endgroup$
    – user549397
    Jan 28 at 0:35
















$begingroup$
$ lambda> 3 $ - are you sure about that part? I think that there should be $ lambda> 2 $
$endgroup$
– VirtualUser
Jan 27 at 22:40




$begingroup$
$ lambda> 3 $ - are you sure about that part? I think that there should be $ lambda> 2 $
$endgroup$
– VirtualUser
Jan 27 at 22:40












$begingroup$
@VirtualUser Yes, since $ cosfrac{1}{h} $ does not exist.
$endgroup$
– user549397
Jan 28 at 0:35




$begingroup$
@VirtualUser Yes, since $ cosfrac{1}{h} $ does not exist.
$endgroup$
– user549397
Jan 28 at 0:35


















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