Proving a question on differentiability

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Here's the question I've been asked to prove:




Suppose that $f$ is differentiable at $c$ and that $f^prime (c)
ne 0$. Show that there exists a $delta > 0$ such that
$0<|x-c|<delta Rightarrow f(x)ne f(c)$.




Here's my attempt:



Assume the contrary that for every $delta > 0$, there is a $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$.



Now, we pick $varepsilon$ such that $0< varepsilon < |f^prime (c)|$. Since $f$ is differentiable at $c$, there is an $delta > 0$ such that if $0<|x-c|<delta$ then we have $left| fracf(x)-f(c)x-c - f^prime (c) right| <epsilon$. For this $delta$, there is $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$. Thus, we must have $left| fracf(x_delta)-f(c)x-c - f^prime (c) right| = left| f^prime (c) right|<epsilon$ which contradicts our choice of $varepsilon$.



Is this proof okay? Is there any possible direct proof? This question is asked in my analysis textbook before "The Mean Value Theorem" section.






real-analysis derivatives







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  • (You can simply use f' for the derivative in $TeX$.)
    – Martin R
    9 mins ago















up vote
1
down vote

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Here's the question I've been asked to prove:




Suppose that $f$ is differentiable at $c$ and that $f^prime (c)
ne 0$. Show that there exists a $delta > 0$ such that
$0<|x-c|<delta Rightarrow f(x)ne f(c)$.




Here's my attempt:



Assume the contrary that for every $delta > 0$, there is a $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$.



Now, we pick $varepsilon$ such that $0< varepsilon < |f^prime (c)|$. Since $f$ is differentiable at $c$, there is an $delta > 0$ such that if $0<|x-c|<delta$ then we have $left| fracf(x)-f(c)x-c - f^prime (c) right| <epsilon$. For this $delta$, there is $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$. Thus, we must have $left| fracf(x_delta)-f(c)x-c - f^prime (c) right| = left| f^prime (c) right|<epsilon$ which contradicts our choice of $varepsilon$.



Is this proof okay? Is there any possible direct proof? This question is asked in my analysis textbook before "The Mean Value Theorem" section.






real-analysis derivatives







share|cite|improve this question





















  • (You can simply use f' for the derivative in $TeX$.)
    – Martin R
    9 mins ago













up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Here's the question I've been asked to prove:




Suppose that $f$ is differentiable at $c$ and that $f^prime (c)
ne 0$. Show that there exists a $delta > 0$ such that
$0<|x-c|<delta Rightarrow f(x)ne f(c)$.




Here's my attempt:



Assume the contrary that for every $delta > 0$, there is a $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$.



Now, we pick $varepsilon$ such that $0< varepsilon < |f^prime (c)|$. Since $f$ is differentiable at $c$, there is an $delta > 0$ such that if $0<|x-c|<delta$ then we have $left| fracf(x)-f(c)x-c - f^prime (c) right| <epsilon$. For this $delta$, there is $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$. Thus, we must have $left| fracf(x_delta)-f(c)x-c - f^prime (c) right| = left| f^prime (c) right|<epsilon$ which contradicts our choice of $varepsilon$.



Is this proof okay? Is there any possible direct proof? This question is asked in my analysis textbook before "The Mean Value Theorem" section.






real-analysis derivatives







share|cite|improve this question













Here's the question I've been asked to prove:




Suppose that $f$ is differentiable at $c$ and that $f^prime (c)
ne 0$. Show that there exists a $delta > 0$ such that
$0<|x-c|<delta Rightarrow f(x)ne f(c)$.




Here's my attempt:



Assume the contrary that for every $delta > 0$, there is a $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$.



Now, we pick $varepsilon$ such that $0< varepsilon < |f^prime (c)|$. Since $f$ is differentiable at $c$, there is an $delta > 0$ such that if $0<|x-c|<delta$ then we have $left| fracf(x)-f(c)x-c - f^prime (c) right| <epsilon$. For this $delta$, there is $x_delta in textdom f$ such that $0<|x_delta-c|<delta$ with $f(x_delta)=f(c)$. Thus, we must have $left| fracf(x_delta)-f(c)x-c - f^prime (c) right| = left| f^prime (c) right|<epsilon$ which contradicts our choice of $varepsilon$.



Is this proof okay? Is there any possible direct proof? This question is asked in my analysis textbook before "The Mean Value Theorem" section.






real-analysis derivatives




real-analysis derivatives






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asked 46 mins ago









Ashish K

596412




596412











  • (You can simply use f' for the derivative in $TeX$.)
    – Martin R
    9 mins ago

















  • (You can simply use f' for the derivative in $TeX$.)
    – Martin R
    9 mins ago
















(You can simply use f' for the derivative in $TeX$.)
– Martin R
9 mins ago





(You can simply use f' for the derivative in $TeX$.)
– Martin R
9 mins ago











3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










You're proof is fine. Here is a bit easier proof in my opinion: suppose that the statement you have to prove is wrong. Then for every $ninmathbbN$ there exists a point $x_n$ such that $0<|x_n-c|<frac1n$ and $f(x_n)=f(c)$. Then we got a sequence that satisfies $cne x_nto c$ but $fracf(x_n)-f(c)x_n-cto 0$. This is a contradiction because by Heine's definition of a limit we must have $fracf(x_n)-f(c)x_n-cto f'(c)$ which is not $0$.






share|cite|improve this answer




















  • Thanks, it's easier using Heine's definition of limit. I wasn't aware of it
    – Ashish K
    31 mins ago






  • 1




    You're solution is also correct, I just prefer to use sequences instead of $epsilon-delta$ when possible. I believe it is a bit easier.
    – Mark
    29 mins ago

















up vote
2
down vote













Here's my initial take on the problem:



If the hypothesis were not true, then look at a sequence of deltas, $delta_n = tfrac1n$. This means there is a sequence of $x_n to c$ as $nto infty$, with $f(x_n) = f(c)$ for all $n geq 1$. Hence
$$lim_nto infty fracf(x_n)-f(c)x_n - c = lim_nto infty 0 = 0,$$
which is a contradiction, since we know that $f'(c)neq 0$ and so any sequence converging to $f'(c)$ should approach this non-zero number.






share|cite|improve this answer



























    up vote
    2
    down vote













    Your proof is fine. With only a small modification you can make it
    a direct argument:



    For $0 < epsilon < |f'(c)|$ there is (as you said) a $delta > 0$ such that
    for $0 < |x - c | < delta$
    $$
    left| fracf(x)-f(c)x-c - f'(c) right| < epsilon , .
    $$
    This implies (using the triangle inequality)
    $$
    left| fracf(x)-f(c)x-c right| > |f'(c)| - epsilon > 0
    $$
    and therefore $f(x) ne c$ for $0 < |x - c | < delta$.






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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      2
      down vote



      accepted










      You're proof is fine. Here is a bit easier proof in my opinion: suppose that the statement you have to prove is wrong. Then for every $ninmathbbN$ there exists a point $x_n$ such that $0<|x_n-c|<frac1n$ and $f(x_n)=f(c)$. Then we got a sequence that satisfies $cne x_nto c$ but $fracf(x_n)-f(c)x_n-cto 0$. This is a contradiction because by Heine's definition of a limit we must have $fracf(x_n)-f(c)x_n-cto f'(c)$ which is not $0$.






      share|cite|improve this answer




















      • Thanks, it's easier using Heine's definition of limit. I wasn't aware of it
        – Ashish K
        31 mins ago






      • 1




        You're solution is also correct, I just prefer to use sequences instead of $epsilon-delta$ when possible. I believe it is a bit easier.
        – Mark
        29 mins ago














      up vote
      2
      down vote



      accepted










      You're proof is fine. Here is a bit easier proof in my opinion: suppose that the statement you have to prove is wrong. Then for every $ninmathbbN$ there exists a point $x_n$ such that $0<|x_n-c|<frac1n$ and $f(x_n)=f(c)$. Then we got a sequence that satisfies $cne x_nto c$ but $fracf(x_n)-f(c)x_n-cto 0$. This is a contradiction because by Heine's definition of a limit we must have $fracf(x_n)-f(c)x_n-cto f'(c)$ which is not $0$.






      share|cite|improve this answer




















      • Thanks, it's easier using Heine's definition of limit. I wasn't aware of it
        – Ashish K
        31 mins ago






      • 1




        You're solution is also correct, I just prefer to use sequences instead of $epsilon-delta$ when possible. I believe it is a bit easier.
        – Mark
        29 mins ago












      up vote
      2
      down vote



      accepted







      up vote
      2
      down vote



      accepted






      You're proof is fine. Here is a bit easier proof in my opinion: suppose that the statement you have to prove is wrong. Then for every $ninmathbbN$ there exists a point $x_n$ such that $0<|x_n-c|<frac1n$ and $f(x_n)=f(c)$. Then we got a sequence that satisfies $cne x_nto c$ but $fracf(x_n)-f(c)x_n-cto 0$. This is a contradiction because by Heine's definition of a limit we must have $fracf(x_n)-f(c)x_n-cto f'(c)$ which is not $0$.






      share|cite|improve this answer












      You're proof is fine. Here is a bit easier proof in my opinion: suppose that the statement you have to prove is wrong. Then for every $ninmathbbN$ there exists a point $x_n$ such that $0<|x_n-c|<frac1n$ and $f(x_n)=f(c)$. Then we got a sequence that satisfies $cne x_nto c$ but $fracf(x_n)-f(c)x_n-cto 0$. This is a contradiction because by Heine's definition of a limit we must have $fracf(x_n)-f(c)x_n-cto f'(c)$ which is not $0$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 38 mins ago









      Mark

      3,024112




      3,024112











      • Thanks, it's easier using Heine's definition of limit. I wasn't aware of it
        – Ashish K
        31 mins ago






      • 1




        You're solution is also correct, I just prefer to use sequences instead of $epsilon-delta$ when possible. I believe it is a bit easier.
        – Mark
        29 mins ago
















      • Thanks, it's easier using Heine's definition of limit. I wasn't aware of it
        – Ashish K
        31 mins ago






      • 1




        You're solution is also correct, I just prefer to use sequences instead of $epsilon-delta$ when possible. I believe it is a bit easier.
        – Mark
        29 mins ago















      Thanks, it's easier using Heine's definition of limit. I wasn't aware of it
      – Ashish K
      31 mins ago




      Thanks, it's easier using Heine's definition of limit. I wasn't aware of it
      – Ashish K
      31 mins ago




      1




      1




      You're solution is also correct, I just prefer to use sequences instead of $epsilon-delta$ when possible. I believe it is a bit easier.
      – Mark
      29 mins ago




      You're solution is also correct, I just prefer to use sequences instead of $epsilon-delta$ when possible. I believe it is a bit easier.
      – Mark
      29 mins ago










      up vote
      2
      down vote













      Here's my initial take on the problem:



      If the hypothesis were not true, then look at a sequence of deltas, $delta_n = tfrac1n$. This means there is a sequence of $x_n to c$ as $nto infty$, with $f(x_n) = f(c)$ for all $n geq 1$. Hence
      $$lim_nto infty fracf(x_n)-f(c)x_n - c = lim_nto infty 0 = 0,$$
      which is a contradiction, since we know that $f'(c)neq 0$ and so any sequence converging to $f'(c)$ should approach this non-zero number.






      share|cite|improve this answer
























        up vote
        2
        down vote













        Here's my initial take on the problem:



        If the hypothesis were not true, then look at a sequence of deltas, $delta_n = tfrac1n$. This means there is a sequence of $x_n to c$ as $nto infty$, with $f(x_n) = f(c)$ for all $n geq 1$. Hence
        $$lim_nto infty fracf(x_n)-f(c)x_n - c = lim_nto infty 0 = 0,$$
        which is a contradiction, since we know that $f'(c)neq 0$ and so any sequence converging to $f'(c)$ should approach this non-zero number.






        share|cite|improve this answer






















          up vote
          2
          down vote










          up vote
          2
          down vote









          Here's my initial take on the problem:



          If the hypothesis were not true, then look at a sequence of deltas, $delta_n = tfrac1n$. This means there is a sequence of $x_n to c$ as $nto infty$, with $f(x_n) = f(c)$ for all $n geq 1$. Hence
          $$lim_nto infty fracf(x_n)-f(c)x_n - c = lim_nto infty 0 = 0,$$
          which is a contradiction, since we know that $f'(c)neq 0$ and so any sequence converging to $f'(c)$ should approach this non-zero number.






          share|cite|improve this answer












          Here's my initial take on the problem:



          If the hypothesis were not true, then look at a sequence of deltas, $delta_n = tfrac1n$. This means there is a sequence of $x_n to c$ as $nto infty$, with $f(x_n) = f(c)$ for all $n geq 1$. Hence
          $$lim_nto infty fracf(x_n)-f(c)x_n - c = lim_nto infty 0 = 0,$$
          which is a contradiction, since we know that $f'(c)neq 0$ and so any sequence converging to $f'(c)$ should approach this non-zero number.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 36 mins ago









          Patch

          1,4931125




          1,4931125




















              up vote
              2
              down vote













              Your proof is fine. With only a small modification you can make it
              a direct argument:



              For $0 < epsilon < |f'(c)|$ there is (as you said) a $delta > 0$ such that
              for $0 < |x - c | < delta$
              $$
              left| fracf(x)-f(c)x-c - f'(c) right| < epsilon , .
              $$
              This implies (using the triangle inequality)
              $$
              left| fracf(x)-f(c)x-c right| > |f'(c)| - epsilon > 0
              $$
              and therefore $f(x) ne c$ for $0 < |x - c | < delta$.






              share|cite|improve this answer


























                up vote
                2
                down vote













                Your proof is fine. With only a small modification you can make it
                a direct argument:



                For $0 < epsilon < |f'(c)|$ there is (as you said) a $delta > 0$ such that
                for $0 < |x - c | < delta$
                $$
                left| fracf(x)-f(c)x-c - f'(c) right| < epsilon , .
                $$
                This implies (using the triangle inequality)
                $$
                left| fracf(x)-f(c)x-c right| > |f'(c)| - epsilon > 0
                $$
                and therefore $f(x) ne c$ for $0 < |x - c | < delta$.






                share|cite|improve this answer
























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Your proof is fine. With only a small modification you can make it
                  a direct argument:



                  For $0 < epsilon < |f'(c)|$ there is (as you said) a $delta > 0$ such that
                  for $0 < |x - c | < delta$
                  $$
                  left| fracf(x)-f(c)x-c - f'(c) right| < epsilon , .
                  $$
                  This implies (using the triangle inequality)
                  $$
                  left| fracf(x)-f(c)x-c right| > |f'(c)| - epsilon > 0
                  $$
                  and therefore $f(x) ne c$ for $0 < |x - c | < delta$.






                  share|cite|improve this answer














                  Your proof is fine. With only a small modification you can make it
                  a direct argument:



                  For $0 < epsilon < |f'(c)|$ there is (as you said) a $delta > 0$ such that
                  for $0 < |x - c | < delta$
                  $$
                  left| fracf(x)-f(c)x-c - f'(c) right| < epsilon , .
                  $$
                  This implies (using the triangle inequality)
                  $$
                  left| fracf(x)-f(c)x-c right| > |f'(c)| - epsilon > 0
                  $$
                  and therefore $f(x) ne c$ for $0 < |x - c | < delta$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 10 mins ago

























                  answered 30 mins ago









                  Martin R

                  24.3k32844




                  24.3k32844



























                       

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