L'Hopital's Rule and subsitution of derivative limit
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
asked yesterday
user2793618
907
add a comment |
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
asked yesterday
user2793618
907
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.
– Lord Shark the Unknown
yesterday
@LordSharktheUnknown Because of the limit laws, specifically for that the limit quotient law.
– user2793618
yesterday
add a comment |
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
asked yesterday
user2793618
907
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
asked yesterday
user2793618
907
asked yesterday
user2793618
907
edited yesterday
asked yesterday
user2793618
907
asked yesterday
user2793618
907
asked yesterday
user2793618
907
907
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.
– Lord Shark the Unknown
yesterday
@LordSharktheUnknown Because of the limit laws, specifically for that the limit quotient law.
– user2793618
yesterday
add a comment |
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.
– Lord Shark the Unknown
yesterday
@LordSharktheUnknown Because of the limit laws, specifically for that the limit quotient law.
– user2793618
yesterday
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.
– Lord Shark the Unknown
yesterday
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.
– Lord Shark the Unknown
yesterday
@LordSharktheUnknown Because of the limit laws, specifically for that the limit quotient law.
– user2793618
yesterday
@LordSharktheUnknown Because of the limit laws, specifically for that the limit quotient law.
– user2793618
yesterday
add a comment |
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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.
– Lord Shark the Unknown
yesterday
@LordSharktheUnknown Because of the limit laws, specifically for that the limit quotient law.
– user2793618
yesterday