L'Hopital's Rule and subsitution of derivative limit












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So we're starting with the limit definition of a derivative and we want to prove something related to this with l'hopital's rule. We'll have to check that $f'$ and $g'$ exist. We're also given that $lim limits_{xto a} f'(x) =k$ exists.



Assuming that limits work 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)}$



After differentiating the numerator, can we just substitute the given $lim limits_{xto a} f'(x) =k$ for $lim limits_{xto a} f'(a)$ ?



Edit: I just realized that it is more likely that we are substituting $lim limits_{xto a} f'(x) =k$ for $lim limits_{xto a} f'(x)$ and that the f'(a) becomes 0 due to differentiation. Is this correct?






calculus proof-writing







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  • $begingroup$
    Why do you assume "limits work like this"? In your second fraction numerator and denominator might both be zero, even though the limit of your first fraction exists.
    $endgroup$
    – Lord Shark the Unknown
    Jan 5 at 7:33










  • $begingroup$
    @LordSharktheUnknown Because of the limit laws, specifically for that the limit quotient law.
    $endgroup$
    – user2793618
    Jan 5 at 7:44
















-1












$begingroup$


So we're starting with the limit definition of a derivative and we want to prove something related to this with l'hopital's rule. We'll have to check that $f'$ and $g'$ exist. We're also given that $lim limits_{xto a} f'(x) =k$ exists.



Assuming that limits work 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)}$



After differentiating the numerator, can we just substitute the given $lim limits_{xto a} f'(x) =k$ for $lim limits_{xto a} f'(a)$ ?



Edit: I just realized that it is more likely that we are substituting $lim limits_{xto a} f'(x) =k$ for $lim limits_{xto a} f'(x)$ and that the f'(a) becomes 0 due to differentiation. Is this correct?






calculus proof-writing







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why do you assume "limits work like this"? In your second fraction numerator and denominator might both be zero, even though the limit of your first fraction exists.
    $endgroup$
    – Lord Shark the Unknown
    Jan 5 at 7:33










  • $begingroup$
    @LordSharktheUnknown Because of the limit laws, specifically for that the limit quotient law.
    $endgroup$
    – user2793618
    Jan 5 at 7:44














-1












-1








-1


1



$begingroup$


So we're starting with the limit definition of a derivative and we want to prove something related to this with l'hopital's rule. We'll have to check that $f'$ and $g'$ exist. We're also given that $lim limits_{xto a} f'(x) =k$ exists.



Assuming that limits work 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)}$



After differentiating the numerator, can we just substitute the given $lim limits_{xto a} f'(x) =k$ for $lim limits_{xto a} f'(a)$ ?



Edit: I just realized that it is more likely that we are substituting $lim limits_{xto a} f'(x) =k$ for $lim limits_{xto a} f'(x)$ and that the f'(a) becomes 0 due to differentiation. Is this correct?






calculus proof-writing







share|cite|improve this question











$endgroup$




So we're starting with the limit definition of a derivative and we want to prove something related to this with l'hopital's rule. We'll have to check that $f'$ and $g'$ exist. We're also given that $lim limits_{xto a} f'(x) =k$ exists.



Assuming that limits work 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)}$



After differentiating the numerator, can we just substitute the given $lim limits_{xto a} f'(x) =k$ for $lim limits_{xto a} f'(a)$ ?



Edit: I just realized that it is more likely that we are substituting $lim limits_{xto a} f'(x) =k$ for $lim limits_{xto a} f'(x)$ and that the f'(a) becomes 0 due to differentiation. Is this correct?






calculus proof-writing




calculus proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 7:48







user2793618

















asked Jan 5 at 7:28









user2793618user2793618

907




907












  • $begingroup$
    Why do you assume "limits work like this"? In your second fraction numerator and denominator might both be zero, even though the limit of your first fraction exists.
    $endgroup$
    – Lord Shark the Unknown
    Jan 5 at 7:33










  • $begingroup$
    @LordSharktheUnknown Because of the limit laws, specifically for that the limit quotient law.
    $endgroup$
    – user2793618
    Jan 5 at 7:44


















  • $begingroup$
    Why do you assume "limits work like this"? In your second fraction numerator and denominator might both be zero, even though the limit of your first fraction exists.
    $endgroup$
    – Lord Shark the Unknown
    Jan 5 at 7:33










  • $begingroup$
    @LordSharktheUnknown Because of the limit laws, specifically for that the limit quotient law.
    $endgroup$
    – user2793618
    Jan 5 at 7:44
















$begingroup$
Why do you assume "limits work like this"? In your second fraction numerator and denominator might both be zero, even though the limit of your first fraction exists.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:33




$begingroup$
Why do you assume "limits work like this"? In your second fraction numerator and denominator might both be zero, even though the limit of your first fraction exists.
$endgroup$
– Lord Shark the Unknown
Jan 5 at 7:33












$begingroup$
@LordSharktheUnknown Because of the limit laws, specifically for that the limit quotient law.
$endgroup$
– user2793618
Jan 5 at 7:44




$begingroup$
@LordSharktheUnknown Because of the limit laws, specifically for that the limit quotient law.
$endgroup$
– user2793618
Jan 5 at 7:44










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