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
asked Jan 5 at 7:28
user2793618user2793618
977
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add a comment |
$begingroup$
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
asked Jan 5 at 7:28
user2793618user2793618
977
$endgroup$
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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.
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– Lord Shark the Unknown
Jan 5 at 7:33
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@LordSharktheUnknown Because of the limit laws, specifically for that the limit quotient law.
$endgroup$
– user2793618
Jan 5 at 7:44
add a comment |
$begingroup$
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
asked Jan 5 at 7:28
user2793618user2793618
977
$endgroup$
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
asked Jan 5 at 7:28
user2793618user2793618
977
asked Jan 5 at 7:28
user2793618user2793618
977
edited Jan 5 at 7:48
user2793618
asked Jan 5 at 7:28
user2793618user2793618
977
asked Jan 5 at 7:28
user2793618user2793618
977
asked Jan 5 at 7:28
user2793618user2793618
977
977
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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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