The Real Operations Behind L'Hopital's Rule












-2












$begingroup$


When using L'Hopital's Rule we have 4 conditions to verify. One of them is that $ lim limits_{xto a} frac{f'(x)}{g'(x)}=c$ where c and a are both real numbers or infinity.



The issue with using L'Hopital's rule is that there are discontinuous derivatives, so when proving something involving L'Hopital's rule it seems to always be good to have $lim limits_{xto a} f'(x) =k$ provided.



Is this due to limits working like this:



$ lim limits_{xto a} frac{f(x)-f(a)}{g(x)-g(a)}=frac{lim limits_{xto a} (f(x)-f(a)) }{lim limits_{xto a} (g(x)-g(a))} = frac{lim limits_{xto a} f(x)-lim limits_{xto a} f(a)}{lim limits_{xto a} g(x)-lim limits_{xto a} g(a)}$



So that when you have the derivative of $f(x)$ (after differentiating the numerator of the example funciton) you can just substitute your assumed $lim limits_{xto a} f'(x) =k$ for $lim limits_{xto a} f'(a)$ ?






calculus limits proof-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What's the actual question here? I don't recall the Hospital requiring the existence of $lim f'(x)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 5 at 6:46










  • $begingroup$
    @LordSharktheUnknown It requires $ lim limits_{xto a} frac{f'(x)}{g'(x)}=c$ to exist. So $f'(x)$ must but in proofs with L'Hopital's rule it seems that having the limit of $f'(x)$ as it approaches a makes proving it possible. I was just wondering if this was due to limit working the way as I hypothesized above. Sorry if the question was unclear.
    $endgroup$
    – user2793618
    Jan 5 at 6:51






  • 1




    $begingroup$
    Look at $lim_{x to infty} frac{f(x)}{g(x)}= lim_{x to infty} frac{sin x}{x^2}$. The limit is obvious but we can still apply LHR to get $lim_{x to infty} frac{f(x)}{g(x)} = lim_{x to infty} frac{f'(x)}{g'(x)} = lim_{x to infty} frac{cos x}{2x} = 0$ even though the limit of the numerator does not exist.
    $endgroup$
    – RRL
    Jan 5 at 7:56










  • $begingroup$
    @RRL isn't that due to the squeeze/sandwich theorem though? And how would you apply the limit definition of derivative to this case? f(a) where a = infinity is undefined. So if you were to make a claim about the derivative of this function (in the limit form) being equal to the limit of the derivative function at infinity, you wouldn't right? You would need the limit of the derivative at infinity?
    $endgroup$
    – user2793618
    Jan 5 at 8:11
















-2












$begingroup$


When using L'Hopital's Rule we have 4 conditions to verify. One of them is that $ lim limits_{xto a} frac{f'(x)}{g'(x)}=c$ where c and a are both real numbers or infinity.



The issue with using L'Hopital's rule is that there are discontinuous derivatives, so when proving something involving L'Hopital's rule it seems to always be good to have $lim limits_{xto a} f'(x) =k$ provided.



Is this due to limits working like this:



$ lim limits_{xto a} frac{f(x)-f(a)}{g(x)-g(a)}=frac{lim limits_{xto a} (f(x)-f(a)) }{lim limits_{xto a} (g(x)-g(a))} = frac{lim limits_{xto a} f(x)-lim limits_{xto a} f(a)}{lim limits_{xto a} g(x)-lim limits_{xto a} g(a)}$



So that when you have the derivative of $f(x)$ (after differentiating the numerator of the example funciton) you can just substitute your assumed $lim limits_{xto a} f'(x) =k$ for $lim limits_{xto a} f'(a)$ ?






calculus limits proof-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What's the actual question here? I don't recall the Hospital requiring the existence of $lim f'(x)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 5 at 6:46










  • $begingroup$
    @LordSharktheUnknown It requires $ lim limits_{xto a} frac{f'(x)}{g'(x)}=c$ to exist. So $f'(x)$ must but in proofs with L'Hopital's rule it seems that having the limit of $f'(x)$ as it approaches a makes proving it possible. I was just wondering if this was due to limit working the way as I hypothesized above. Sorry if the question was unclear.
    $endgroup$
    – user2793618
    Jan 5 at 6:51






  • 1




    $begingroup$
    Look at $lim_{x to infty} frac{f(x)}{g(x)}= lim_{x to infty} frac{sin x}{x^2}$. The limit is obvious but we can still apply LHR to get $lim_{x to infty} frac{f(x)}{g(x)} = lim_{x to infty} frac{f'(x)}{g'(x)} = lim_{x to infty} frac{cos x}{2x} = 0$ even though the limit of the numerator does not exist.
    $endgroup$
    – RRL
    Jan 5 at 7:56










  • $begingroup$
    @RRL isn't that due to the squeeze/sandwich theorem though? And how would you apply the limit definition of derivative to this case? f(a) where a = infinity is undefined. So if you were to make a claim about the derivative of this function (in the limit form) being equal to the limit of the derivative function at infinity, you wouldn't right? You would need the limit of the derivative at infinity?
    $endgroup$
    – user2793618
    Jan 5 at 8:11














-2












-2








-2





$begingroup$


When using L'Hopital's Rule we have 4 conditions to verify. One of them is that $ lim limits_{xto a} frac{f'(x)}{g'(x)}=c$ where c and a are both real numbers or infinity.



The issue with using L'Hopital's rule is that there are discontinuous derivatives, so when proving something involving L'Hopital's rule it seems to always be good to have $lim limits_{xto a} f'(x) =k$ provided.



Is this due to limits working like this:



$ lim limits_{xto a} frac{f(x)-f(a)}{g(x)-g(a)}=frac{lim limits_{xto a} (f(x)-f(a)) }{lim limits_{xto a} (g(x)-g(a))} = frac{lim limits_{xto a} f(x)-lim limits_{xto a} f(a)}{lim limits_{xto a} g(x)-lim limits_{xto a} g(a)}$



So that when you have the derivative of $f(x)$ (after differentiating the numerator of the example funciton) you can just substitute your assumed $lim limits_{xto a} f'(x) =k$ for $lim limits_{xto a} f'(a)$ ?






calculus limits proof-theory







share|cite|improve this question











$endgroup$




When using L'Hopital's Rule we have 4 conditions to verify. One of them is that $ lim limits_{xto a} frac{f'(x)}{g'(x)}=c$ where c and a are both real numbers or infinity.



The issue with using L'Hopital's rule is that there are discontinuous derivatives, so when proving something involving L'Hopital's rule it seems to always be good to have $lim limits_{xto a} f'(x) =k$ provided.



Is this due to limits working like this:



$ lim limits_{xto a} frac{f(x)-f(a)}{g(x)-g(a)}=frac{lim limits_{xto a} (f(x)-f(a)) }{lim limits_{xto a} (g(x)-g(a))} = frac{lim limits_{xto a} f(x)-lim limits_{xto a} f(a)}{lim limits_{xto a} g(x)-lim limits_{xto a} g(a)}$



So that when you have the derivative of $f(x)$ (after differentiating the numerator of the example funciton) you can just substitute your assumed $lim limits_{xto a} f'(x) =k$ for $lim limits_{xto a} f'(a)$ ?






calculus limits proof-theory




calculus limits proof-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 7:16







user2793618

















asked Jan 5 at 6:44









user2793618user2793618

977




977








  • 1




    $begingroup$
    What's the actual question here? I don't recall the Hospital requiring the existence of $lim f'(x)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 5 at 6:46










  • $begingroup$
    @LordSharktheUnknown It requires $ lim limits_{xto a} frac{f'(x)}{g'(x)}=c$ to exist. So $f'(x)$ must but in proofs with L'Hopital's rule it seems that having the limit of $f'(x)$ as it approaches a makes proving it possible. I was just wondering if this was due to limit working the way as I hypothesized above. Sorry if the question was unclear.
    $endgroup$
    – user2793618
    Jan 5 at 6:51






  • 1




    $begingroup$
    Look at $lim_{x to infty} frac{f(x)}{g(x)}= lim_{x to infty} frac{sin x}{x^2}$. The limit is obvious but we can still apply LHR to get $lim_{x to infty} frac{f(x)}{g(x)} = lim_{x to infty} frac{f'(x)}{g'(x)} = lim_{x to infty} frac{cos x}{2x} = 0$ even though the limit of the numerator does not exist.
    $endgroup$
    – RRL
    Jan 5 at 7:56










  • $begingroup$
    @RRL isn't that due to the squeeze/sandwich theorem though? And how would you apply the limit definition of derivative to this case? f(a) where a = infinity is undefined. So if you were to make a claim about the derivative of this function (in the limit form) being equal to the limit of the derivative function at infinity, you wouldn't right? You would need the limit of the derivative at infinity?
    $endgroup$
    – user2793618
    Jan 5 at 8:11














  • 1




    $begingroup$
    What's the actual question here? I don't recall the Hospital requiring the existence of $lim f'(x)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 5 at 6:46










  • $begingroup$
    @LordSharktheUnknown It requires $ lim limits_{xto a} frac{f'(x)}{g'(x)}=c$ to exist. So $f'(x)$ must but in proofs with L'Hopital's rule it seems that having the limit of $f'(x)$ as it approaches a makes proving it possible. I was just wondering if this was due to limit working the way as I hypothesized above. Sorry if the question was unclear.
    $endgroup$
    – user2793618
    Jan 5 at 6:51






  • 1




    $begingroup$
    Look at $lim_{x to infty} frac{f(x)}{g(x)}= lim_{x to infty} frac{sin x}{x^2}$. The limit is obvious but we can still apply LHR to get $lim_{x to infty} frac{f(x)}{g(x)} = lim_{x to infty} frac{f'(x)}{g'(x)} = lim_{x to infty} frac{cos x}{2x} = 0$ even though the limit of the numerator does not exist.
    $endgroup$
    – RRL
    Jan 5 at 7:56










  • $begingroup$
    @RRL isn't that due to the squeeze/sandwich theorem though? And how would you apply the limit definition of derivative to this case? f(a) where a = infinity is undefined. So if you were to make a claim about the derivative of this function (in the limit form) being equal to the limit of the derivative function at infinity, you wouldn't right? You would need the limit of the derivative at infinity?
    $endgroup$
    – user2793618
    Jan 5 at 8:11








1




1




$begingroup$
What's the actual question here? I don't recall the Hospital requiring the existence of $lim f'(x)$.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 6:46




$begingroup$
What's the actual question here? I don't recall the Hospital requiring the existence of $lim f'(x)$.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 6:46












$begingroup$
@LordSharktheUnknown It requires $ lim limits_{xto a} frac{f'(x)}{g'(x)}=c$ to exist. So $f'(x)$ must but in proofs with L'Hopital's rule it seems that having the limit of $f'(x)$ as it approaches a makes proving it possible. I was just wondering if this was due to limit working the way as I hypothesized above. Sorry if the question was unclear.
$endgroup$
– user2793618
Jan 5 at 6:51




$begingroup$
@LordSharktheUnknown It requires $ lim limits_{xto a} frac{f'(x)}{g'(x)}=c$ to exist. So $f'(x)$ must but in proofs with L'Hopital's rule it seems that having the limit of $f'(x)$ as it approaches a makes proving it possible. I was just wondering if this was due to limit working the way as I hypothesized above. Sorry if the question was unclear.
$endgroup$
– user2793618
Jan 5 at 6:51




1




1




$begingroup$
Look at $lim_{x to infty} frac{f(x)}{g(x)}= lim_{x to infty} frac{sin x}{x^2}$. The limit is obvious but we can still apply LHR to get $lim_{x to infty} frac{f(x)}{g(x)} = lim_{x to infty} frac{f'(x)}{g'(x)} = lim_{x to infty} frac{cos x}{2x} = 0$ even though the limit of the numerator does not exist.
$endgroup$
– RRL
Jan 5 at 7:56




$begingroup$
Look at $lim_{x to infty} frac{f(x)}{g(x)}= lim_{x to infty} frac{sin x}{x^2}$. The limit is obvious but we can still apply LHR to get $lim_{x to infty} frac{f(x)}{g(x)} = lim_{x to infty} frac{f'(x)}{g'(x)} = lim_{x to infty} frac{cos x}{2x} = 0$ even though the limit of the numerator does not exist.
$endgroup$
– RRL
Jan 5 at 7:56












$begingroup$
@RRL isn't that due to the squeeze/sandwich theorem though? And how would you apply the limit definition of derivative to this case? f(a) where a = infinity is undefined. So if you were to make a claim about the derivative of this function (in the limit form) being equal to the limit of the derivative function at infinity, you wouldn't right? You would need the limit of the derivative at infinity?
$endgroup$
– user2793618
Jan 5 at 8:11




$begingroup$
@RRL isn't that due to the squeeze/sandwich theorem though? And how would you apply the limit definition of derivative to this case? f(a) where a = infinity is undefined. So if you were to make a claim about the derivative of this function (in the limit form) being equal to the limit of the derivative function at infinity, you wouldn't right? You would need the limit of the derivative at infinity?
$endgroup$
– user2793618
Jan 5 at 8:11










1 Answer
1






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0












$begingroup$

L'hospital's rule works because of this: If $f$ and $g$ are differentiable at $x=a$ then by linear approximation:



$$f(x) approx f(a) + f'(a)(x-a)$$



and



$$g(x) approx g(a) + g'(a)(x-a).$$



Since we're assuming $f(a)=g(a) =0$ we have



$$ frac{f(x)}{g(x)} approx
frac{f(a) + f'(a)(x-a)}{g(a) + g'(a)(x-a)}
= frac{0 + f'(a)(x-a)}{0 + g'(a)(x-a)} = frac{f'(a)}{g'(a)}.$$






share|cite|improve this answer









$endgroup$













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    active

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    0












    $begingroup$

    L'hospital's rule works because of this: If $f$ and $g$ are differentiable at $x=a$ then by linear approximation:



    $$f(x) approx f(a) + f'(a)(x-a)$$



    and



    $$g(x) approx g(a) + g'(a)(x-a).$$



    Since we're assuming $f(a)=g(a) =0$ we have



    $$ frac{f(x)}{g(x)} approx
    frac{f(a) + f'(a)(x-a)}{g(a) + g'(a)(x-a)}
    = frac{0 + f'(a)(x-a)}{0 + g'(a)(x-a)} = frac{f'(a)}{g'(a)}.$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      L'hospital's rule works because of this: If $f$ and $g$ are differentiable at $x=a$ then by linear approximation:



      $$f(x) approx f(a) + f'(a)(x-a)$$



      and



      $$g(x) approx g(a) + g'(a)(x-a).$$



      Since we're assuming $f(a)=g(a) =0$ we have



      $$ frac{f(x)}{g(x)} approx
      frac{f(a) + f'(a)(x-a)}{g(a) + g'(a)(x-a)}
      = frac{0 + f'(a)(x-a)}{0 + g'(a)(x-a)} = frac{f'(a)}{g'(a)}.$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        L'hospital's rule works because of this: If $f$ and $g$ are differentiable at $x=a$ then by linear approximation:



        $$f(x) approx f(a) + f'(a)(x-a)$$



        and



        $$g(x) approx g(a) + g'(a)(x-a).$$



        Since we're assuming $f(a)=g(a) =0$ we have



        $$ frac{f(x)}{g(x)} approx
        frac{f(a) + f'(a)(x-a)}{g(a) + g'(a)(x-a)}
        = frac{0 + f'(a)(x-a)}{0 + g'(a)(x-a)} = frac{f'(a)}{g'(a)}.$$






        share|cite|improve this answer









        $endgroup$



        L'hospital's rule works because of this: If $f$ and $g$ are differentiable at $x=a$ then by linear approximation:



        $$f(x) approx f(a) + f'(a)(x-a)$$



        and



        $$g(x) approx g(a) + g'(a)(x-a).$$



        Since we're assuming $f(a)=g(a) =0$ we have



        $$ frac{f(x)}{g(x)} approx
        frac{f(a) + f'(a)(x-a)}{g(a) + g'(a)(x-a)}
        = frac{0 + f'(a)(x-a)}{0 + g'(a)(x-a)} = frac{f'(a)}{g'(a)}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 10:30









        B. GoddardB. Goddard

        18.9k21440




        18.9k21440






























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