Limit of a derivative of function at infinity
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If we have a differentiable function $f:(0,∞) rightarrow mathbb{R}$ such that derivative $f'(t)$ tends to zero as $t rightarrow ∞$ and for $n in mathbb{Z}, f(n) rightarrow l$ (finite) as $n rightarrow ∞$, then how do we show that $f(t) rightarrow l$ as $t (in mathbb{R}) rightarrow ∞$.
What I don't understand is that if for $n in mathbb{Z}, f(n) rightarrow l$ as $n rightarrow ∞$, then shouldn't this hold for any real number $t$ as well? We can always find an integer greater than any real number. What is the use of limit of derivative in solving this question?
sequences-and-series limits analysis derivatives
asked Dec 2 '18 at 20:42
SALONI SINHASALONI SINHA
162
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add a comment |
$begingroup$
If we have a differentiable function $f:(0,∞) rightarrow mathbb{R}$ such that derivative $f'(t)$ tends to zero as $t rightarrow ∞$ and for $n in mathbb{Z}, f(n) rightarrow l$ (finite) as $n rightarrow ∞$, then how do we show that $f(t) rightarrow l$ as $t (in mathbb{R}) rightarrow ∞$.
What I don't understand is that if for $n in mathbb{Z}, f(n) rightarrow l$ as $n rightarrow ∞$, then shouldn't this hold for any real number $t$ as well? We can always find an integer greater than any real number. What is the use of limit of derivative in solving this question?
sequences-and-series limits analysis derivatives
asked Dec 2 '18 at 20:42
SALONI SINHASALONI SINHA
162
$endgroup$
add a comment |
$begingroup$
If we have a differentiable function $f:(0,∞) rightarrow mathbb{R}$ such that derivative $f'(t)$ tends to zero as $t rightarrow ∞$ and for $n in mathbb{Z}, f(n) rightarrow l$ (finite) as $n rightarrow ∞$, then how do we show that $f(t) rightarrow l$ as $t (in mathbb{R}) rightarrow ∞$.
What I don't understand is that if for $n in mathbb{Z}, f(n) rightarrow l$ as $n rightarrow ∞$, then shouldn't this hold for any real number $t$ as well? We can always find an integer greater than any real number. What is the use of limit of derivative in solving this question?
sequences-and-series limits analysis derivatives
asked Dec 2 '18 at 20:42
SALONI SINHASALONI SINHA
162
$endgroup$
If we have a differentiable function $f:(0,∞) rightarrow mathbb{R}$ such that derivative $f'(t)$ tends to zero as $t rightarrow ∞$ and for $n in mathbb{Z}, f(n) rightarrow l$ (finite) as $n rightarrow ∞$, then how do we show that $f(t) rightarrow l$ as $t (in mathbb{R}) rightarrow ∞$.
What I don't understand is that if for $n in mathbb{Z}, f(n) rightarrow l$ as $n rightarrow ∞$, then shouldn't this hold for any real number $t$ as well? We can always find an integer greater than any real number. What is the use of limit of derivative in solving this question?
sequences-and-series limits analysis derivatives
sequences-and-series limits analysis derivatives
asked Dec 2 '18 at 20:42
SALONI SINHASALONI SINHA
162
asked Dec 2 '18 at 20:42
SALONI SINHASALONI SINHA
162
asked Dec 2 '18 at 20:42
SALONI SINHASALONI SINHA
162
asked Dec 2 '18 at 20:42
SALONI SINHASALONI SINHA
162
asked Dec 2 '18 at 20:42
SALONI SINHASALONI SINHA
162
162
add a comment |
add a comment |
2 Answers
2
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oldest
votes
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Take the function $f(t) = sin (pi t)$ and $l=0$. Then $f(n) = 0$ for all $n$ but $f(n+{ 1 over 2}) = 1$ for all $n$.
You need to have a condition that limits the behaviour of $f$ on $(n,n+1)$ for
large $n$. This is what the $f'(t) to 0$ condition does.
answered Dec 2 '18 at 20:53
copper.hatcopper.hat
127k559160
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add a comment |
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We have that $forall xin[n,n+1]$ by MVT
$$frac{f(x)-f(n)}{x-n}=f'(c) quad cin(n,n+1) implies |f(x)-f(n)|=(x-n)|f'(c)|le |f'(c)|to 0$$
answered Dec 2 '18 at 20:46
gimusigimusi
92.8k84494
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take the function $f(t) = sin (pi t)$ and $l=0$. Then $f(n) = 0$ for all $n$ but $f(n+{ 1 over 2}) = 1$ for all $n$.
You need to have a condition that limits the behaviour of $f$ on $(n,n+1)$ for
large $n$. This is what the $f'(t) to 0$ condition does.
answered Dec 2 '18 at 20:53
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
$begingroup$
Take the function $f(t) = sin (pi t)$ and $l=0$. Then $f(n) = 0$ for all $n$ but $f(n+{ 1 over 2}) = 1$ for all $n$.
You need to have a condition that limits the behaviour of $f$ on $(n,n+1)$ for
large $n$. This is what the $f'(t) to 0$ condition does.
answered Dec 2 '18 at 20:53
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
$begingroup$
Take the function $f(t) = sin (pi t)$ and $l=0$. Then $f(n) = 0$ for all $n$ but $f(n+{ 1 over 2}) = 1$ for all $n$.
You need to have a condition that limits the behaviour of $f$ on $(n,n+1)$ for
large $n$. This is what the $f'(t) to 0$ condition does.
answered Dec 2 '18 at 20:53
copper.hatcopper.hat
127k559160
$endgroup$
Take the function $f(t) = sin (pi t)$ and $l=0$. Then $f(n) = 0$ for all $n$ but $f(n+{ 1 over 2}) = 1$ for all $n$.
You need to have a condition that limits the behaviour of $f$ on $(n,n+1)$ for
large $n$. This is what the $f'(t) to 0$ condition does.
answered Dec 2 '18 at 20:53
copper.hatcopper.hat
127k559160
answered Dec 2 '18 at 20:53
copper.hatcopper.hat
127k559160
answered Dec 2 '18 at 20:53
copper.hatcopper.hat
127k559160
answered Dec 2 '18 at 20:53
copper.hatcopper.hat
127k559160
127k559160
add a comment |
add a comment |
$begingroup$
We have that $forall xin[n,n+1]$ by MVT
$$frac{f(x)-f(n)}{x-n}=f'(c) quad cin(n,n+1) implies |f(x)-f(n)|=(x-n)|f'(c)|le |f'(c)|to 0$$
answered Dec 2 '18 at 20:46
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
We have that $forall xin[n,n+1]$ by MVT
$$frac{f(x)-f(n)}{x-n}=f'(c) quad cin(n,n+1) implies |f(x)-f(n)|=(x-n)|f'(c)|le |f'(c)|to 0$$
answered Dec 2 '18 at 20:46
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
We have that $forall xin[n,n+1]$ by MVT
$$frac{f(x)-f(n)}{x-n}=f'(c) quad cin(n,n+1) implies |f(x)-f(n)|=(x-n)|f'(c)|le |f'(c)|to 0$$
answered Dec 2 '18 at 20:46
gimusigimusi
92.8k84494
$endgroup$
We have that $forall xin[n,n+1]$ by MVT
$$frac{f(x)-f(n)}{x-n}=f'(c) quad cin(n,n+1) implies |f(x)-f(n)|=(x-n)|f'(c)|le |f'(c)|to 0$$
answered Dec 2 '18 at 20:46
gimusigimusi
92.8k84494
answered Dec 2 '18 at 20:46
gimusigimusi
92.8k84494
answered Dec 2 '18 at 20:46
gimusigimusi
92.8k84494
answered Dec 2 '18 at 20:46
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
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