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Prove $f:mathbbR^nto mathbbR$ if differentiable is continuous
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Below I have written a proof for the problem in my title and was wondering if the proof works
$$0=lim_hto 0|f(a+h)-f(a)-triangledown fcdot h|$$
$$=lim_hto 0|f(a+h)-f(a)-big(f_xh_1+cdots f_nh_nbig)|$$
and since $h to 0$ implies $(h_1,cdots ,h_n) Longrightarrow (0,cdots , 0)$
$$big(f_xh_1+cdots f_nh_nbig) =0 text as hto 0$$
So we get continuity since
$$0=lim_hto 0|f(a+h)-f(a)|$$
calculus real-analysis proof-verification
asked Jul 23 at 0:24
john fowles
1,088817
add a comment |Â
up vote
0
down vote
favorite
Below I have written a proof for the problem in my title and was wondering if the proof works
$$0=lim_hto 0|f(a+h)-f(a)-triangledown fcdot h|$$
$$=lim_hto 0|f(a+h)-f(a)-big(f_xh_1+cdots f_nh_nbig)|$$
and since $h to 0$ implies $(h_1,cdots ,h_n) Longrightarrow (0,cdots , 0)$
$$big(f_xh_1+cdots f_nh_nbig) =0 text as hto 0$$
So we get continuity since
$$0=lim_hto 0|f(a+h)-f(a)|$$
calculus real-analysis proof-verification
asked Jul 23 at 0:24
john fowles
1,088817
In one dimension suppose $f(a^+)ne f(a^-)$, but $ lim_hto 0fracf(a^++h)-f(a^+)h= lim_hto 0fracf(a^-)-f(a^--h)h$. Does this define the derivative at $a$?
– herb steinberg
Jul 23 at 0:31
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Below I have written a proof for the problem in my title and was wondering if the proof works
$$0=lim_hto 0|f(a+h)-f(a)-triangledown fcdot h|$$
$$=lim_hto 0|f(a+h)-f(a)-big(f_xh_1+cdots f_nh_nbig)|$$
and since $h to 0$ implies $(h_1,cdots ,h_n) Longrightarrow (0,cdots , 0)$
$$big(f_xh_1+cdots f_nh_nbig) =0 text as hto 0$$
So we get continuity since
$$0=lim_hto 0|f(a+h)-f(a)|$$
calculus real-analysis proof-verification
asked Jul 23 at 0:24
john fowles
1,088817
Below I have written a proof for the problem in my title and was wondering if the proof works
$$0=lim_hto 0|f(a+h)-f(a)-triangledown fcdot h|$$
$$=lim_hto 0|f(a+h)-f(a)-big(f_xh_1+cdots f_nh_nbig)|$$
and since $h to 0$ implies $(h_1,cdots ,h_n) Longrightarrow (0,cdots , 0)$
$$big(f_xh_1+cdots f_nh_nbig) =0 text as hto 0$$
So we get continuity since
$$0=lim_hto 0|f(a+h)-f(a)|$$
calculus real-analysis proof-verification
asked Jul 23 at 0:24
john fowles
1,088817
edited Jul 23 at 0:29
asked Jul 23 at 0:24
john fowles
1,088817
asked Jul 23 at 0:24
john fowles
1,088817
asked Jul 23 at 0:24
john fowles
1,088817
1,088817
In one dimension suppose $f(a^+)ne f(a^-)$, but $ lim_hto 0fracf(a^++h)-f(a^+)h= lim_hto 0fracf(a^-)-f(a^--h)h$. Does this define the derivative at $a$?
– herb steinberg
Jul 23 at 0:31
add a comment |Â
In one dimension suppose $f(a^+)ne f(a^-)$, but $ lim_hto 0fracf(a^++h)-f(a^+)h= lim_hto 0fracf(a^-)-f(a^--h)h$. Does this define the derivative at $a$?
– herb steinberg
Jul 23 at 0:31
In one dimension suppose $f(a^+)ne f(a^-)$, but $ lim_hto 0fracf(a^++h)-f(a^+)h= lim_hto 0fracf(a^-)-f(a^--h)h$. Does this define the derivative at $a$?
– herb steinberg
Jul 23 at 0:31
In one dimension suppose $f(a^+)ne f(a^-)$, but $ lim_hto 0fracf(a^++h)-f(a^+)h= lim_hto 0fracf(a^-)-f(a^--h)h$. Does this define the derivative at $a$?
– herb steinberg
Jul 23 at 0:31
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Here's a proof from the book (also one in English :)).
beginalign* lim_x to x_0 f(x) = f(x_0) &iff lim_x to x_0 |f(x)- f(x_0)| = 0 \ \ &iff lim_x to x_0 left|f(x)- f(x_0)right| fracx-x_0x-x_0= 0 \ \ & iff lim_x to x_0 fracx-x_0|x-x_0| = 0 \ \ & iff lim_x to x_0 left(fracx-x_0right)|x-x_0| = 0 \ \ & leq lim_x to x_0 left(frac + x-x_0right)|x-x_0| \ \ & = lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 |x-x_0| + |f'(x_0)(x-x_0)| endalign*
By the differentiability assumption we have,
$$lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 = 0$$
Hence, each limit exists separately i.e;
beginalign* &lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 |x-x_0| + |f'(x_0)(x-x_0)| \ \ & =lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 |x-x_0| + lim_x to x_0 |f'(x_0)(x-x_0)| \ \ & = 0 endalign*
Therefore,
$$ lim_x to x_0 |f(x)-f(x_0)| = 0 Rightarrow lim_x to x_0 f(x) = f(x_0)$$
answered Jul 23 at 1:04
Faraad Armwood
7,4292619
This is clear. Thanks. and just wondering, were there any problems with my proof?
– john fowles
Jul 23 at 4:51
Yes, when you write that second equality with the limit, the quantities on each side of the minus sign don't go to zero with different orders and so you can just call $nabla f(a) cdot h$ zero in the limit and keep the other term i.e $f(a+h) - f(a)$
– Faraad Armwood
Jul 23 at 5:02
What do you mean by "go to zero with different orders". I assume you mean that if I had $|2h|-|h|$ then these terms approach $0$ at different rates where $2h$ will equal $0$ faster? But more importantly, is what your hinting at a rule that states that I can't get rid of one term based on that reason if it applies to more terms in the expression?
– john fowles
Jul 23 at 5:23
Yes and yes. Sorry about the confusion.
– Faraad Armwood
Jul 23 at 15:06
add a comment |Â
up vote
1
down vote
Probar que si $f:mathbbR^nlongrightarrow mathbbR^n$ es diferenciable en $ainmathbbR^n$, entonces es continua en $a$.
Supongamos que $f$ es diferenciable en $ainmathbbR^n$, es decir, existe una única transformación lineal $lambda:mathbbR^nlongrightarrowmathbbR^n$ tal que
$$
displaystylelim_hlongrightarrow 0 fracvert f(a+h)-f(a)-lambda (h)vertvert hvert=0.
$$
Observe que
$$
beginarraycccccc
0 & leq & vert f(a+h)-f(a)vert & = & left vert lambda(h)+vert hvert cdot dfracf(a+h)-f(a)-lambda (h)vert hvertrightvert \
& & & leq & vertlambda (h)vert +vert hvertcdotdfracvert f(a+h)-f(a)-lambda (h)vertvert hvert \
& & & leq & Mcdotvert hvert +vert hvertcdotdfracvert f(a+h)-f(a)-lambda (h)vertvert hvert
endarray
$$
AsÃÂ,
$$
displaystylelim_hlongrightarrow 0 0leqdisplaystylelim_hlongrightarrow 0 vert f(a+h)-f(a)vertleqdisplaystylelim_hlongrightarrow 0 vert hvertleft(vertlambdavert+dfracvert f(a+h)-f(a)-lambda (h)vertvert hvertright)
$$
$
beginarrayrl
Rightarrow & displaystylelim_hlongrightarrow 0 vert f(a+h)-f(a)vert=0 \
& x=a+h qquad hlongrightarrow 0 Leftrightarrow xlongrightarrow a \
Rightarrow & displaystylelim_xlongrightarrow a vert f(x)-f(a)vert=0 \
Rightarrow & displaystylelim_xlongrightarrow a f(x)=f(a) \
Rightarrow & f mbox es continua en a
endarray
$
answered Jul 23 at 0:49
Moisés Rojas
111
I know you speak spanish, but he's asking in english hahaha
– Gonzalo Benavides
Jul 23 at 4:20
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Here's a proof from the book (also one in English :)).
beginalign* lim_x to x_0 f(x) = f(x_0) &iff lim_x to x_0 |f(x)- f(x_0)| = 0 \ \ &iff lim_x to x_0 left|f(x)- f(x_0)right| fracx-x_0x-x_0= 0 \ \ & iff lim_x to x_0 fracx-x_0|x-x_0| = 0 \ \ & iff lim_x to x_0 left(fracx-x_0right)|x-x_0| = 0 \ \ & leq lim_x to x_0 left(frac + x-x_0right)|x-x_0| \ \ & = lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 |x-x_0| + |f'(x_0)(x-x_0)| endalign*
By the differentiability assumption we have,
$$lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 = 0$$
Hence, each limit exists separately i.e;
beginalign* &lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 |x-x_0| + |f'(x_0)(x-x_0)| \ \ & =lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 |x-x_0| + lim_x to x_0 |f'(x_0)(x-x_0)| \ \ & = 0 endalign*
Therefore,
$$ lim_x to x_0 |f(x)-f(x_0)| = 0 Rightarrow lim_x to x_0 f(x) = f(x_0)$$
answered Jul 23 at 1:04
Faraad Armwood
7,4292619
This is clear. Thanks. and just wondering, were there any problems with my proof?
– john fowles
Jul 23 at 4:51
Yes, when you write that second equality with the limit, the quantities on each side of the minus sign don't go to zero with different orders and so you can just call $nabla f(a) cdot h$ zero in the limit and keep the other term i.e $f(a+h) - f(a)$
– Faraad Armwood
Jul 23 at 5:02
What do you mean by "go to zero with different orders". I assume you mean that if I had $|2h|-|h|$ then these terms approach $0$ at different rates where $2h$ will equal $0$ faster? But more importantly, is what your hinting at a rule that states that I can't get rid of one term based on that reason if it applies to more terms in the expression?
– john fowles
Jul 23 at 5:23
Yes and yes. Sorry about the confusion.
– Faraad Armwood
Jul 23 at 15:06
add a comment |Â
up vote
1
down vote
accepted
Here's a proof from the book (also one in English :)).
beginalign* lim_x to x_0 f(x) = f(x_0) &iff lim_x to x_0 |f(x)- f(x_0)| = 0 \ \ &iff lim_x to x_0 left|f(x)- f(x_0)right| fracx-x_0x-x_0= 0 \ \ & iff lim_x to x_0 fracx-x_0|x-x_0| = 0 \ \ & iff lim_x to x_0 left(fracx-x_0right)|x-x_0| = 0 \ \ & leq lim_x to x_0 left(frac + x-x_0right)|x-x_0| \ \ & = lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 |x-x_0| + |f'(x_0)(x-x_0)| endalign*
By the differentiability assumption we have,
$$lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 = 0$$
Hence, each limit exists separately i.e;
beginalign* &lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 |x-x_0| + |f'(x_0)(x-x_0)| \ \ & =lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 |x-x_0| + lim_x to x_0 |f'(x_0)(x-x_0)| \ \ & = 0 endalign*
Therefore,
$$ lim_x to x_0 |f(x)-f(x_0)| = 0 Rightarrow lim_x to x_0 f(x) = f(x_0)$$
answered Jul 23 at 1:04
Faraad Armwood
7,4292619
This is clear. Thanks. and just wondering, were there any problems with my proof?
– john fowles
Jul 23 at 4:51
Yes, when you write that second equality with the limit, the quantities on each side of the minus sign don't go to zero with different orders and so you can just call $nabla f(a) cdot h$ zero in the limit and keep the other term i.e $f(a+h) - f(a)$
– Faraad Armwood
Jul 23 at 5:02
What do you mean by "go to zero with different orders". I assume you mean that if I had $|2h|-|h|$ then these terms approach $0$ at different rates where $2h$ will equal $0$ faster? But more importantly, is what your hinting at a rule that states that I can't get rid of one term based on that reason if it applies to more terms in the expression?
– john fowles
Jul 23 at 5:23
Yes and yes. Sorry about the confusion.
– Faraad Armwood
Jul 23 at 15:06
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Here's a proof from the book (also one in English :)).
beginalign* lim_x to x_0 f(x) = f(x_0) &iff lim_x to x_0 |f(x)- f(x_0)| = 0 \ \ &iff lim_x to x_0 left|f(x)- f(x_0)right| fracx-x_0x-x_0= 0 \ \ & iff lim_x to x_0 fracx-x_0|x-x_0| = 0 \ \ & iff lim_x to x_0 left(fracx-x_0right)|x-x_0| = 0 \ \ & leq lim_x to x_0 left(frac + x-x_0right)|x-x_0| \ \ & = lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 |x-x_0| + |f'(x_0)(x-x_0)| endalign*
By the differentiability assumption we have,
$$lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 = 0$$
Hence, each limit exists separately i.e;
beginalign* &lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 |x-x_0| + |f'(x_0)(x-x_0)| \ \ & =lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 |x-x_0| + lim_x to x_0 |f'(x_0)(x-x_0)| \ \ & = 0 endalign*
Therefore,
$$ lim_x to x_0 |f(x)-f(x_0)| = 0 Rightarrow lim_x to x_0 f(x) = f(x_0)$$
answered Jul 23 at 1:04
Faraad Armwood
7,4292619
Here's a proof from the book (also one in English :)).
beginalign* lim_x to x_0 f(x) = f(x_0) &iff lim_x to x_0 |f(x)- f(x_0)| = 0 \ \ &iff lim_x to x_0 left|f(x)- f(x_0)right| fracx-x_0x-x_0= 0 \ \ & iff lim_x to x_0 fracx-x_0|x-x_0| = 0 \ \ & iff lim_x to x_0 left(fracx-x_0right)|x-x_0| = 0 \ \ & leq lim_x to x_0 left(frac + x-x_0right)|x-x_0| \ \ & = lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 |x-x_0| + |f'(x_0)(x-x_0)| endalign*
By the differentiability assumption we have,
$$lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 = 0$$
Hence, each limit exists separately i.e;
beginalign* &lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 |x-x_0| + |f'(x_0)(x-x_0)| \ \ & =lim_x to x_0 fracf(x)-f(x_0)-f'(x_0)(x-x_0)x-x_0 |x-x_0| + lim_x to x_0 |f'(x_0)(x-x_0)| \ \ & = 0 endalign*
Therefore,
$$ lim_x to x_0 |f(x)-f(x_0)| = 0 Rightarrow lim_x to x_0 f(x) = f(x_0)$$
answered Jul 23 at 1:04
Faraad Armwood
7,4292619
edited Jul 23 at 1:30
answered Jul 23 at 1:04
Faraad Armwood
7,4292619
answered Jul 23 at 1:04
Faraad Armwood
7,4292619
answered Jul 23 at 1:04
Faraad Armwood
7,4292619
7,4292619
This is clear. Thanks. and just wondering, were there any problems with my proof?
– john fowles
Jul 23 at 4:51
Yes, when you write that second equality with the limit, the quantities on each side of the minus sign don't go to zero with different orders and so you can just call $nabla f(a) cdot h$ zero in the limit and keep the other term i.e $f(a+h) - f(a)$
– Faraad Armwood
Jul 23 at 5:02
What do you mean by "go to zero with different orders". I assume you mean that if I had $|2h|-|h|$ then these terms approach $0$ at different rates where $2h$ will equal $0$ faster? But more importantly, is what your hinting at a rule that states that I can't get rid of one term based on that reason if it applies to more terms in the expression?
– john fowles
Jul 23 at 5:23
Yes and yes. Sorry about the confusion.
– Faraad Armwood
Jul 23 at 15:06
add a comment |Â
This is clear. Thanks. and just wondering, were there any problems with my proof?
– john fowles
Jul 23 at 4:51
Yes, when you write that second equality with the limit, the quantities on each side of the minus sign don't go to zero with different orders and so you can just call $nabla f(a) cdot h$ zero in the limit and keep the other term i.e $f(a+h) - f(a)$
– Faraad Armwood
Jul 23 at 5:02
What do you mean by "go to zero with different orders". I assume you mean that if I had $|2h|-|h|$ then these terms approach $0$ at different rates where $2h$ will equal $0$ faster? But more importantly, is what your hinting at a rule that states that I can't get rid of one term based on that reason if it applies to more terms in the expression?
– john fowles
Jul 23 at 5:23
Yes and yes. Sorry about the confusion.
– Faraad Armwood
Jul 23 at 15:06
This is clear. Thanks. and just wondering, were there any problems with my proof?
– john fowles
Jul 23 at 4:51
This is clear. Thanks. and just wondering, were there any problems with my proof?
– john fowles
Jul 23 at 4:51
Yes, when you write that second equality with the limit, the quantities on each side of the minus sign don't go to zero with different orders and so you can just call $nabla f(a) cdot h$ zero in the limit and keep the other term i.e $f(a+h) - f(a)$
– Faraad Armwood
Jul 23 at 5:02
Yes, when you write that second equality with the limit, the quantities on each side of the minus sign don't go to zero with different orders and so you can just call $nabla f(a) cdot h$ zero in the limit and keep the other term i.e $f(a+h) - f(a)$
– Faraad Armwood
Jul 23 at 5:02
What do you mean by "go to zero with different orders". I assume you mean that if I had $|2h|-|h|$ then these terms approach $0$ at different rates where $2h$ will equal $0$ faster? But more importantly, is what your hinting at a rule that states that I can't get rid of one term based on that reason if it applies to more terms in the expression?
– john fowles
Jul 23 at 5:23
What do you mean by "go to zero with different orders". I assume you mean that if I had $|2h|-|h|$ then these terms approach $0$ at different rates where $2h$ will equal $0$ faster? But more importantly, is what your hinting at a rule that states that I can't get rid of one term based on that reason if it applies to more terms in the expression?
– john fowles
Jul 23 at 5:23
Yes and yes. Sorry about the confusion.
– Faraad Armwood
Jul 23 at 15:06
Yes and yes. Sorry about the confusion.
– Faraad Armwood
Jul 23 at 15:06
add a comment |Â
up vote
1
down vote
Probar que si $f:mathbbR^nlongrightarrow mathbbR^n$ es diferenciable en $ainmathbbR^n$, entonces es continua en $a$.
Supongamos que $f$ es diferenciable en $ainmathbbR^n$, es decir, existe una única transformación lineal $lambda:mathbbR^nlongrightarrowmathbbR^n$ tal que
$$
displaystylelim_hlongrightarrow 0 fracvert f(a+h)-f(a)-lambda (h)vertvert hvert=0.
$$
Observe que
$$
beginarraycccccc
0 & leq & vert f(a+h)-f(a)vert & = & left vert lambda(h)+vert hvert cdot dfracf(a+h)-f(a)-lambda (h)vert hvertrightvert \
& & & leq & vertlambda (h)vert +vert hvertcdotdfracvert f(a+h)-f(a)-lambda (h)vertvert hvert \
& & & leq & Mcdotvert hvert +vert hvertcdotdfracvert f(a+h)-f(a)-lambda (h)vertvert hvert
endarray
$$
AsÃÂ,
$$
displaystylelim_hlongrightarrow 0 0leqdisplaystylelim_hlongrightarrow 0 vert f(a+h)-f(a)vertleqdisplaystylelim_hlongrightarrow 0 vert hvertleft(vertlambdavert+dfracvert f(a+h)-f(a)-lambda (h)vertvert hvertright)
$$
$
beginarrayrl
Rightarrow & displaystylelim_hlongrightarrow 0 vert f(a+h)-f(a)vert=0 \
& x=a+h qquad hlongrightarrow 0 Leftrightarrow xlongrightarrow a \
Rightarrow & displaystylelim_xlongrightarrow a vert f(x)-f(a)vert=0 \
Rightarrow & displaystylelim_xlongrightarrow a f(x)=f(a) \
Rightarrow & f mbox es continua en a
endarray
$
answered Jul 23 at 0:49
Moisés Rojas
111
I know you speak spanish, but he's asking in english hahaha
– Gonzalo Benavides
Jul 23 at 4:20
add a comment |Â
up vote
1
down vote
Probar que si $f:mathbbR^nlongrightarrow mathbbR^n$ es diferenciable en $ainmathbbR^n$, entonces es continua en $a$.
Supongamos que $f$ es diferenciable en $ainmathbbR^n$, es decir, existe una única transformación lineal $lambda:mathbbR^nlongrightarrowmathbbR^n$ tal que
$$
displaystylelim_hlongrightarrow 0 fracvert f(a+h)-f(a)-lambda (h)vertvert hvert=0.
$$
Observe que
$$
beginarraycccccc
0 & leq & vert f(a+h)-f(a)vert & = & left vert lambda(h)+vert hvert cdot dfracf(a+h)-f(a)-lambda (h)vert hvertrightvert \
& & & leq & vertlambda (h)vert +vert hvertcdotdfracvert f(a+h)-f(a)-lambda (h)vertvert hvert \
& & & leq & Mcdotvert hvert +vert hvertcdotdfracvert f(a+h)-f(a)-lambda (h)vertvert hvert
endarray
$$
AsÃÂ,
$$
displaystylelim_hlongrightarrow 0 0leqdisplaystylelim_hlongrightarrow 0 vert f(a+h)-f(a)vertleqdisplaystylelim_hlongrightarrow 0 vert hvertleft(vertlambdavert+dfracvert f(a+h)-f(a)-lambda (h)vertvert hvertright)
$$
$
beginarrayrl
Rightarrow & displaystylelim_hlongrightarrow 0 vert f(a+h)-f(a)vert=0 \
& x=a+h qquad hlongrightarrow 0 Leftrightarrow xlongrightarrow a \
Rightarrow & displaystylelim_xlongrightarrow a vert f(x)-f(a)vert=0 \
Rightarrow & displaystylelim_xlongrightarrow a f(x)=f(a) \
Rightarrow & f mbox es continua en a
endarray
$
answered Jul 23 at 0:49
Moisés Rojas
111
I know you speak spanish, but he's asking in english hahaha
– Gonzalo Benavides
Jul 23 at 4:20
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Probar que si $f:mathbbR^nlongrightarrow mathbbR^n$ es diferenciable en $ainmathbbR^n$, entonces es continua en $a$.
Supongamos que $f$ es diferenciable en $ainmathbbR^n$, es decir, existe una única transformación lineal $lambda:mathbbR^nlongrightarrowmathbbR^n$ tal que
$$
displaystylelim_hlongrightarrow 0 fracvert f(a+h)-f(a)-lambda (h)vertvert hvert=0.
$$
Observe que
$$
beginarraycccccc
0 & leq & vert f(a+h)-f(a)vert & = & left vert lambda(h)+vert hvert cdot dfracf(a+h)-f(a)-lambda (h)vert hvertrightvert \
& & & leq & vertlambda (h)vert +vert hvertcdotdfracvert f(a+h)-f(a)-lambda (h)vertvert hvert \
& & & leq & Mcdotvert hvert +vert hvertcdotdfracvert f(a+h)-f(a)-lambda (h)vertvert hvert
endarray
$$
AsÃÂ,
$$
displaystylelim_hlongrightarrow 0 0leqdisplaystylelim_hlongrightarrow 0 vert f(a+h)-f(a)vertleqdisplaystylelim_hlongrightarrow 0 vert hvertleft(vertlambdavert+dfracvert f(a+h)-f(a)-lambda (h)vertvert hvertright)
$$
$
beginarrayrl
Rightarrow & displaystylelim_hlongrightarrow 0 vert f(a+h)-f(a)vert=0 \
& x=a+h qquad hlongrightarrow 0 Leftrightarrow xlongrightarrow a \
Rightarrow & displaystylelim_xlongrightarrow a vert f(x)-f(a)vert=0 \
Rightarrow & displaystylelim_xlongrightarrow a f(x)=f(a) \
Rightarrow & f mbox es continua en a
endarray
$
answered Jul 23 at 0:49
Moisés Rojas
111
Probar que si $f:mathbbR^nlongrightarrow mathbbR^n$ es diferenciable en $ainmathbbR^n$, entonces es continua en $a$.
Supongamos que $f$ es diferenciable en $ainmathbbR^n$, es decir, existe una única transformación lineal $lambda:mathbbR^nlongrightarrowmathbbR^n$ tal que
$$
displaystylelim_hlongrightarrow 0 fracvert f(a+h)-f(a)-lambda (h)vertvert hvert=0.
$$
Observe que
$$
beginarraycccccc
0 & leq & vert f(a+h)-f(a)vert & = & left vert lambda(h)+vert hvert cdot dfracf(a+h)-f(a)-lambda (h)vert hvertrightvert \
& & & leq & vertlambda (h)vert +vert hvertcdotdfracvert f(a+h)-f(a)-lambda (h)vertvert hvert \
& & & leq & Mcdotvert hvert +vert hvertcdotdfracvert f(a+h)-f(a)-lambda (h)vertvert hvert
endarray
$$
AsÃÂ,
$$
displaystylelim_hlongrightarrow 0 0leqdisplaystylelim_hlongrightarrow 0 vert f(a+h)-f(a)vertleqdisplaystylelim_hlongrightarrow 0 vert hvertleft(vertlambdavert+dfracvert f(a+h)-f(a)-lambda (h)vertvert hvertright)
$$
$
beginarrayrl
Rightarrow & displaystylelim_hlongrightarrow 0 vert f(a+h)-f(a)vert=0 \
& x=a+h qquad hlongrightarrow 0 Leftrightarrow xlongrightarrow a \
Rightarrow & displaystylelim_xlongrightarrow a vert f(x)-f(a)vert=0 \
Rightarrow & displaystylelim_xlongrightarrow a f(x)=f(a) \
Rightarrow & f mbox es continua en a
endarray
$
answered Jul 23 at 0:49
Moisés Rojas
111
answered Jul 23 at 0:49
Moisés Rojas
111
answered Jul 23 at 0:49
Moisés Rojas
111
answered Jul 23 at 0:49
Moisés Rojas
111
111
I know you speak spanish, but he's asking in english hahaha
– Gonzalo Benavides
Jul 23 at 4:20
add a comment |Â
I know you speak spanish, but he's asking in english hahaha
– Gonzalo Benavides
Jul 23 at 4:20
I know you speak spanish, but he's asking in english hahaha
– Gonzalo Benavides
Jul 23 at 4:20
I know you speak spanish, but he's asking in english hahaha
– Gonzalo Benavides
Jul 23 at 4:20
add a comment |Â
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In one dimension suppose $f(a^+)ne f(a^-)$, but $ lim_hto 0fracf(a^++h)-f(a^+)h= lim_hto 0fracf(a^-)-f(a^--h)h$. Does this define the derivative at $a$?
– herb steinberg
Jul 23 at 0:31