Mixing
![Sum of series $1 - 1/3 + 1/6 - 1/10 + ldots$? [duplicate]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Continuity proof for exponential
Clash Royale CLAN TAG#URR8PPP
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0
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Show that $f(x) = e^x$ is continuous using the epsilon-delta definition.
I can't seem to write down anything meaningful...
real-analysis continuity exponential-function
edited Mar 23 '14 at 7:10
Martin Sleziak
43.5k6113259
asked Mar 23 '14 at 0:32
kiwifruit
299319
 |Â
show 4 more comments
up vote
0
down vote
favorite
Show that $f(x) = e^x$ is continuous using the epsilon-delta definition.
I can't seem to write down anything meaningful...
real-analysis continuity exponential-function
edited Mar 23 '14 at 7:10
Martin Sleziak
43.5k6113259
asked Mar 23 '14 at 0:32
kiwifruit
299319
1
What is known about $e^x$, and how is it defined?
– Daniel Fischer♦
Mar 23 '14 at 0:33
I can see that is is continuous geometrically of course, but not sure how to get this formally
– kiwifruit
Mar 23 '14 at 0:34
1
How to prove it formally depends on what you already know, and which definition you have. The power series definition, $$e^x = sum_n=0^infty fracx^nn!,?$$ Do you have the addition theorem, $e^x+y = e^xcdot e^y$?
– Daniel Fischer♦
Mar 23 '14 at 0:36
1
Apply the definition of continuity everywhere: for all $c$ and for all $epsilon > 0$, there exists $delta > 0$ such that if $|x - c| < delta$, then $|f(x) - f(c)| < epsilon$. So since $e^x$ is defined everywhere on $mathbbR$, let $c in mathbbR$ and let $epsilon > 0$. Can you produce a corresponding $delta > 0$?
– ml0105
Mar 23 '14 at 0:38
2
$lvert e^x-e^crvert = e^ccdotlvert e^x-c-1rvert$. That reduces the problem to showing continuity in $0$.
– Daniel Fischer♦
Mar 23 '14 at 0:42
 |Â
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Show that $f(x) = e^x$ is continuous using the epsilon-delta definition.
I can't seem to write down anything meaningful...
real-analysis continuity exponential-function
edited Mar 23 '14 at 7:10
Martin Sleziak
43.5k6113259
asked Mar 23 '14 at 0:32
kiwifruit
299319
Show that $f(x) = e^x$ is continuous using the epsilon-delta definition.
I can't seem to write down anything meaningful...
real-analysis continuity exponential-function
edited Mar 23 '14 at 7:10
Martin Sleziak
43.5k6113259
asked Mar 23 '14 at 0:32
kiwifruit
299319
edited Mar 23 '14 at 7:10
Martin Sleziak
43.5k6113259
edited Mar 23 '14 at 7:10
Martin Sleziak
43.5k6113259
edited Mar 23 '14 at 7:10
Martin Sleziak
43.5k6113259
43.5k6113259
asked Mar 23 '14 at 0:32
kiwifruit
299319
asked Mar 23 '14 at 0:32
kiwifruit
299319
asked Mar 23 '14 at 0:32
kiwifruit
299319
299319
1
What is known about $e^x$, and how is it defined?
– Daniel Fischer♦
Mar 23 '14 at 0:33
I can see that is is continuous geometrically of course, but not sure how to get this formally
– kiwifruit
Mar 23 '14 at 0:34
1
How to prove it formally depends on what you already know, and which definition you have. The power series definition, $$e^x = sum_n=0^infty fracx^nn!,?$$ Do you have the addition theorem, $e^x+y = e^xcdot e^y$?
– Daniel Fischer♦
Mar 23 '14 at 0:36
1
Apply the definition of continuity everywhere: for all $c$ and for all $epsilon > 0$, there exists $delta > 0$ such that if $|x - c| < delta$, then $|f(x) - f(c)| < epsilon$. So since $e^x$ is defined everywhere on $mathbbR$, let $c in mathbbR$ and let $epsilon > 0$. Can you produce a corresponding $delta > 0$?
– ml0105
Mar 23 '14 at 0:38
2
$lvert e^x-e^crvert = e^ccdotlvert e^x-c-1rvert$. That reduces the problem to showing continuity in $0$.
– Daniel Fischer♦
Mar 23 '14 at 0:42
 |Â
show 4 more comments
1
What is known about $e^x$, and how is it defined?
– Daniel Fischer♦
Mar 23 '14 at 0:33
I can see that is is continuous geometrically of course, but not sure how to get this formally
– kiwifruit
Mar 23 '14 at 0:34
1
How to prove it formally depends on what you already know, and which definition you have. The power series definition, $$e^x = sum_n=0^infty fracx^nn!,?$$ Do you have the addition theorem, $e^x+y = e^xcdot e^y$?
– Daniel Fischer♦
Mar 23 '14 at 0:36
1
Apply the definition of continuity everywhere: for all $c$ and for all $epsilon > 0$, there exists $delta > 0$ such that if $|x - c| < delta$, then $|f(x) - f(c)| < epsilon$. So since $e^x$ is defined everywhere on $mathbbR$, let $c in mathbbR$ and let $epsilon > 0$. Can you produce a corresponding $delta > 0$?
– ml0105
Mar 23 '14 at 0:38
2
$lvert e^x-e^crvert = e^ccdotlvert e^x-c-1rvert$. That reduces the problem to showing continuity in $0$.
– Daniel Fischer♦
Mar 23 '14 at 0:42
1
1
What is known about $e^x$, and how is it defined?
– Daniel Fischer♦
Mar 23 '14 at 0:33
What is known about $e^x$, and how is it defined?
– Daniel Fischer♦
Mar 23 '14 at 0:33
I can see that is is continuous geometrically of course, but not sure how to get this formally
– kiwifruit
Mar 23 '14 at 0:34
I can see that is is continuous geometrically of course, but not sure how to get this formally
– kiwifruit
Mar 23 '14 at 0:34
1
1
How to prove it formally depends on what you already know, and which definition you have. The power series definition, $$e^x = sum_n=0^infty fracx^nn!,?$$ Do you have the addition theorem, $e^x+y = e^xcdot e^y$?
– Daniel Fischer♦
Mar 23 '14 at 0:36
How to prove it formally depends on what you already know, and which definition you have. The power series definition, $$e^x = sum_n=0^infty fracx^nn!,?$$ Do you have the addition theorem, $e^x+y = e^xcdot e^y$?
– Daniel Fischer♦
Mar 23 '14 at 0:36
1
1
Apply the definition of continuity everywhere: for all $c$ and for all $epsilon > 0$, there exists $delta > 0$ such that if $|x - c| < delta$, then $|f(x) - f(c)| < epsilon$. So since $e^x$ is defined everywhere on $mathbbR$, let $c in mathbbR$ and let $epsilon > 0$. Can you produce a corresponding $delta > 0$?
– ml0105
Mar 23 '14 at 0:38
Apply the definition of continuity everywhere: for all $c$ and for all $epsilon > 0$, there exists $delta > 0$ such that if $|x - c| < delta$, then $|f(x) - f(c)| < epsilon$. So since $e^x$ is defined everywhere on $mathbbR$, let $c in mathbbR$ and let $epsilon > 0$. Can you produce a corresponding $delta > 0$?
– ml0105
Mar 23 '14 at 0:38
2
2
$lvert e^x-e^crvert = e^ccdotlvert e^x-c-1rvert$. That reduces the problem to showing continuity in $0$.
– Daniel Fischer♦
Mar 23 '14 at 0:42
$lvert e^x-e^crvert = e^ccdotlvert e^x-c-1rvert$. That reduces the problem to showing continuity in $0$.
– Daniel Fischer♦
Mar 23 '14 at 0:42
 |Â
show 4 more comments
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Almost identical questions have been answered $n + 1$ times on this site. The proof is trivial. I will present a slight variation of the standard proof.
Recall that, by definition, $f(x)$ is continuous on $mathbbR$ if and only if
$$(forall x_0 in mathbbR)(forall varepsilon > 0)(exists delta > 0)(forall x in mathbbR)(0 < |x - x_0| < delta implies |f(x) - f(x_0)| < varepsilon).$$
Combining a well-worn result
$$e^x = lim_n to inftyleft(1 + fracxnright)^n,$$
and the fact that, by definition, $lim_x to x_0 f(x) = L$ if and only if
$$(forall varepsilon > 0)(exists delta > 0)(0 < |x - x_0| < delta implies |f(x) - L| < varepsilon),$$
proves that $e^x$ is continuous.
Edit: To summarize, we know that the sequence
$$s_n = left(1 + fracxnright)^n$$
converges to $e^x$ as $n to infty$. We see that
$$lim_x to x_0 lim_n to infty left(1 + fracxnright)^n = lim_x to x_0 e^x = e^x_0 = lim_n to infty left(1 + fracx_0nright)^n = lim_n to infty lim_x to x_0 left(1 + fracxnright)^n.$$
Moreover, by definition, $lim_x to x_0 f(x) = L$ if and only if
$$(forall varepsilon > 0)(exists delta > 0)(0 < |x - x_0| < delta implies |f(x) - L| < varepsilon).$$
Yet, since $x_0$ is arbitrary, we have
$$(forall x_0 in mathbbR)(forall varepsilon > 0)(exists delta > 0)(forall x in mathbbR)(0 < |x - x_0| < delta implies |f(x) - f(x_0)| < varepsilon).$$
Thus, $e^x$ is continuous for all $x$.
answered Mar 23 '14 at 0:50
glebovg
6,85222044
Who downvoted my answer? Please elaborate.
– glebovg
Mar 23 '14 at 1:05
2
I didn't downvote you, but I think I'm missing something. Using your outline, one would need to show that $lim_x to x_0 e^x = e^x_0$, or $lim_x to x_0 lim_n to infty (1 + fracxn)^n = lim_n to infty (1 + fracx_0n)^n$, which amounts to justifying interchanging the limits, no?
– anonymous
Mar 23 '14 at 4:39
@anonymous Yes. I edited my answer to address your comment.
– glebovg
Mar 23 '14 at 7:06
Oh, ha, I see. Cute!
– anonymous
Mar 23 '14 at 15:40
add a comment |Â
up vote
3
down vote
Let $a$ be a positive real number. Then the function $f: mathbbRto mathbbR$ defined by $xmapsto a^x$ is continuous.
Proof:
1) First we prove the continuous at $0$.
We may assume that $a>1$. Let $varepsilon>0$ be arbitrary. Since $a^1/kto 1$ and $a^-1/kto 1$ as $kto infty$, we choose $K$ such that both are $varepsilon$-close to $1$. Let $delta=1/K$, so for $|x|<1/K$ we have
beginalign-1/K<x<1/K\a^-1/K<a^x<a^1/K endalign
which proves that $a^x$ is $varepsilon$-close to $1$ as desired (the case $ale1$ is handled similarly just with the inequality in the other direction).
2) To conclude we prove the continuity in the general case.
Let $x_0in mathbbR$, we have to show that $lim_xto x_0a^x=a^x_0$. Since $x-x_0to 0$ as $xto x_0$, then $a^x-x_0to1$ by ($1$) and thus
$$lim_xto x_0a^x=lim_xto x_0a^x_0a^x-x_0=a^x_0lim_xto x_0a^x-x_0=a^x_0$$
as was to be shown.
Since $ein mathbbR^>0$, thus $e^x$ is continuous.
answered Mar 23 '14 at 5:38
Jose Antonio
4,36921326
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up vote
2
down vote
Start with $|x-c|<delta$.
Let's take a look at the case that $xge c$.
Then:
$$x-c <delta$$
$$x < c+ delta$$
$$e^x < e^c+delta$$
$$0 le e^x-e^c < e^c(e^delta - 1)$$
Let's pick $delta$ such that $varepsilon = e^c(e^delta - 1)$.
Then:
$$delta = ln(1+varepsilon e^-c)$$
Repeat for $x<c$.
In other words, you are able to find a $delta$ for any $varepsilon>0$.
Qed.
answered Mar 23 '14 at 0:47
I like Serena
3,0731718
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Almost identical questions have been answered $n + 1$ times on this site. The proof is trivial. I will present a slight variation of the standard proof.
Recall that, by definition, $f(x)$ is continuous on $mathbbR$ if and only if
$$(forall x_0 in mathbbR)(forall varepsilon > 0)(exists delta > 0)(forall x in mathbbR)(0 < |x - x_0| < delta implies |f(x) - f(x_0)| < varepsilon).$$
Combining a well-worn result
$$e^x = lim_n to inftyleft(1 + fracxnright)^n,$$
and the fact that, by definition, $lim_x to x_0 f(x) = L$ if and only if
$$(forall varepsilon > 0)(exists delta > 0)(0 < |x - x_0| < delta implies |f(x) - L| < varepsilon),$$
proves that $e^x$ is continuous.
Edit: To summarize, we know that the sequence
$$s_n = left(1 + fracxnright)^n$$
converges to $e^x$ as $n to infty$. We see that
$$lim_x to x_0 lim_n to infty left(1 + fracxnright)^n = lim_x to x_0 e^x = e^x_0 = lim_n to infty left(1 + fracx_0nright)^n = lim_n to infty lim_x to x_0 left(1 + fracxnright)^n.$$
Moreover, by definition, $lim_x to x_0 f(x) = L$ if and only if
$$(forall varepsilon > 0)(exists delta > 0)(0 < |x - x_0| < delta implies |f(x) - L| < varepsilon).$$
Yet, since $x_0$ is arbitrary, we have
$$(forall x_0 in mathbbR)(forall varepsilon > 0)(exists delta > 0)(forall x in mathbbR)(0 < |x - x_0| < delta implies |f(x) - f(x_0)| < varepsilon).$$
Thus, $e^x$ is continuous for all $x$.
answered Mar 23 '14 at 0:50
glebovg
6,85222044
Who downvoted my answer? Please elaborate.
– glebovg
Mar 23 '14 at 1:05
2
I didn't downvote you, but I think I'm missing something. Using your outline, one would need to show that $lim_x to x_0 e^x = e^x_0$, or $lim_x to x_0 lim_n to infty (1 + fracxn)^n = lim_n to infty (1 + fracx_0n)^n$, which amounts to justifying interchanging the limits, no?
– anonymous
Mar 23 '14 at 4:39
@anonymous Yes. I edited my answer to address your comment.
– glebovg
Mar 23 '14 at 7:06
Oh, ha, I see. Cute!
– anonymous
Mar 23 '14 at 15:40
add a comment |Â
up vote
2
down vote
accepted
Almost identical questions have been answered $n + 1$ times on this site. The proof is trivial. I will present a slight variation of the standard proof.
Recall that, by definition, $f(x)$ is continuous on $mathbbR$ if and only if
$$(forall x_0 in mathbbR)(forall varepsilon > 0)(exists delta > 0)(forall x in mathbbR)(0 < |x - x_0| < delta implies |f(x) - f(x_0)| < varepsilon).$$
Combining a well-worn result
$$e^x = lim_n to inftyleft(1 + fracxnright)^n,$$
and the fact that, by definition, $lim_x to x_0 f(x) = L$ if and only if
$$(forall varepsilon > 0)(exists delta > 0)(0 < |x - x_0| < delta implies |f(x) - L| < varepsilon),$$
proves that $e^x$ is continuous.
Edit: To summarize, we know that the sequence
$$s_n = left(1 + fracxnright)^n$$
converges to $e^x$ as $n to infty$. We see that
$$lim_x to x_0 lim_n to infty left(1 + fracxnright)^n = lim_x to x_0 e^x = e^x_0 = lim_n to infty left(1 + fracx_0nright)^n = lim_n to infty lim_x to x_0 left(1 + fracxnright)^n.$$
Moreover, by definition, $lim_x to x_0 f(x) = L$ if and only if
$$(forall varepsilon > 0)(exists delta > 0)(0 < |x - x_0| < delta implies |f(x) - L| < varepsilon).$$
Yet, since $x_0$ is arbitrary, we have
$$(forall x_0 in mathbbR)(forall varepsilon > 0)(exists delta > 0)(forall x in mathbbR)(0 < |x - x_0| < delta implies |f(x) - f(x_0)| < varepsilon).$$
Thus, $e^x$ is continuous for all $x$.
answered Mar 23 '14 at 0:50
glebovg
6,85222044
Who downvoted my answer? Please elaborate.
– glebovg
Mar 23 '14 at 1:05
2
I didn't downvote you, but I think I'm missing something. Using your outline, one would need to show that $lim_x to x_0 e^x = e^x_0$, or $lim_x to x_0 lim_n to infty (1 + fracxn)^n = lim_n to infty (1 + fracx_0n)^n$, which amounts to justifying interchanging the limits, no?
– anonymous
Mar 23 '14 at 4:39
@anonymous Yes. I edited my answer to address your comment.
– glebovg
Mar 23 '14 at 7:06
Oh, ha, I see. Cute!
– anonymous
Mar 23 '14 at 15:40
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Almost identical questions have been answered $n + 1$ times on this site. The proof is trivial. I will present a slight variation of the standard proof.
Recall that, by definition, $f(x)$ is continuous on $mathbbR$ if and only if
$$(forall x_0 in mathbbR)(forall varepsilon > 0)(exists delta > 0)(forall x in mathbbR)(0 < |x - x_0| < delta implies |f(x) - f(x_0)| < varepsilon).$$
Combining a well-worn result
$$e^x = lim_n to inftyleft(1 + fracxnright)^n,$$
and the fact that, by definition, $lim_x to x_0 f(x) = L$ if and only if
$$(forall varepsilon > 0)(exists delta > 0)(0 < |x - x_0| < delta implies |f(x) - L| < varepsilon),$$
proves that $e^x$ is continuous.
Edit: To summarize, we know that the sequence
$$s_n = left(1 + fracxnright)^n$$
converges to $e^x$ as $n to infty$. We see that
$$lim_x to x_0 lim_n to infty left(1 + fracxnright)^n = lim_x to x_0 e^x = e^x_0 = lim_n to infty left(1 + fracx_0nright)^n = lim_n to infty lim_x to x_0 left(1 + fracxnright)^n.$$
Moreover, by definition, $lim_x to x_0 f(x) = L$ if and only if
$$(forall varepsilon > 0)(exists delta > 0)(0 < |x - x_0| < delta implies |f(x) - L| < varepsilon).$$
Yet, since $x_0$ is arbitrary, we have
$$(forall x_0 in mathbbR)(forall varepsilon > 0)(exists delta > 0)(forall x in mathbbR)(0 < |x - x_0| < delta implies |f(x) - f(x_0)| < varepsilon).$$
Thus, $e^x$ is continuous for all $x$.
answered Mar 23 '14 at 0:50
glebovg
6,85222044
Almost identical questions have been answered $n + 1$ times on this site. The proof is trivial. I will present a slight variation of the standard proof.
Recall that, by definition, $f(x)$ is continuous on $mathbbR$ if and only if
$$(forall x_0 in mathbbR)(forall varepsilon > 0)(exists delta > 0)(forall x in mathbbR)(0 < |x - x_0| < delta implies |f(x) - f(x_0)| < varepsilon).$$
Combining a well-worn result
$$e^x = lim_n to inftyleft(1 + fracxnright)^n,$$
and the fact that, by definition, $lim_x to x_0 f(x) = L$ if and only if
$$(forall varepsilon > 0)(exists delta > 0)(0 < |x - x_0| < delta implies |f(x) - L| < varepsilon),$$
proves that $e^x$ is continuous.
Edit: To summarize, we know that the sequence
$$s_n = left(1 + fracxnright)^n$$
converges to $e^x$ as $n to infty$. We see that
$$lim_x to x_0 lim_n to infty left(1 + fracxnright)^n = lim_x to x_0 e^x = e^x_0 = lim_n to infty left(1 + fracx_0nright)^n = lim_n to infty lim_x to x_0 left(1 + fracxnright)^n.$$
Moreover, by definition, $lim_x to x_0 f(x) = L$ if and only if
$$(forall varepsilon > 0)(exists delta > 0)(0 < |x - x_0| < delta implies |f(x) - L| < varepsilon).$$
Yet, since $x_0$ is arbitrary, we have
$$(forall x_0 in mathbbR)(forall varepsilon > 0)(exists delta > 0)(forall x in mathbbR)(0 < |x - x_0| < delta implies |f(x) - f(x_0)| < varepsilon).$$
Thus, $e^x$ is continuous for all $x$.
answered Mar 23 '14 at 0:50
glebovg
6,85222044
edited Mar 23 '14 at 7:20
answered Mar 23 '14 at 0:50
glebovg
6,85222044
answered Mar 23 '14 at 0:50
glebovg
6,85222044
answered Mar 23 '14 at 0:50
glebovg
6,85222044
6,85222044
Who downvoted my answer? Please elaborate.
– glebovg
Mar 23 '14 at 1:05
2
I didn't downvote you, but I think I'm missing something. Using your outline, one would need to show that $lim_x to x_0 e^x = e^x_0$, or $lim_x to x_0 lim_n to infty (1 + fracxn)^n = lim_n to infty (1 + fracx_0n)^n$, which amounts to justifying interchanging the limits, no?
– anonymous
Mar 23 '14 at 4:39
@anonymous Yes. I edited my answer to address your comment.
– glebovg
Mar 23 '14 at 7:06
Oh, ha, I see. Cute!
– anonymous
Mar 23 '14 at 15:40
add a comment |Â
Who downvoted my answer? Please elaborate.
– glebovg
Mar 23 '14 at 1:05
2
I didn't downvote you, but I think I'm missing something. Using your outline, one would need to show that $lim_x to x_0 e^x = e^x_0$, or $lim_x to x_0 lim_n to infty (1 + fracxn)^n = lim_n to infty (1 + fracx_0n)^n$, which amounts to justifying interchanging the limits, no?
– anonymous
Mar 23 '14 at 4:39
@anonymous Yes. I edited my answer to address your comment.
– glebovg
Mar 23 '14 at 7:06
Oh, ha, I see. Cute!
– anonymous
Mar 23 '14 at 15:40
Who downvoted my answer? Please elaborate.
– glebovg
Mar 23 '14 at 1:05
Who downvoted my answer? Please elaborate.
– glebovg
Mar 23 '14 at 1:05
2
2
I didn't downvote you, but I think I'm missing something. Using your outline, one would need to show that $lim_x to x_0 e^x = e^x_0$, or $lim_x to x_0 lim_n to infty (1 + fracxn)^n = lim_n to infty (1 + fracx_0n)^n$, which amounts to justifying interchanging the limits, no?
– anonymous
Mar 23 '14 at 4:39
I didn't downvote you, but I think I'm missing something. Using your outline, one would need to show that $lim_x to x_0 e^x = e^x_0$, or $lim_x to x_0 lim_n to infty (1 + fracxn)^n = lim_n to infty (1 + fracx_0n)^n$, which amounts to justifying interchanging the limits, no?
– anonymous
Mar 23 '14 at 4:39
@anonymous Yes. I edited my answer to address your comment.
– glebovg
Mar 23 '14 at 7:06
@anonymous Yes. I edited my answer to address your comment.
– glebovg
Mar 23 '14 at 7:06
Oh, ha, I see. Cute!
– anonymous
Mar 23 '14 at 15:40
Oh, ha, I see. Cute!
– anonymous
Mar 23 '14 at 15:40
add a comment |Â
up vote
3
down vote
Let $a$ be a positive real number. Then the function $f: mathbbRto mathbbR$ defined by $xmapsto a^x$ is continuous.
Proof:
1) First we prove the continuous at $0$.
We may assume that $a>1$. Let $varepsilon>0$ be arbitrary. Since $a^1/kto 1$ and $a^-1/kto 1$ as $kto infty$, we choose $K$ such that both are $varepsilon$-close to $1$. Let $delta=1/K$, so for $|x|<1/K$ we have
beginalign-1/K<x<1/K\a^-1/K<a^x<a^1/K endalign
which proves that $a^x$ is $varepsilon$-close to $1$ as desired (the case $ale1$ is handled similarly just with the inequality in the other direction).
2) To conclude we prove the continuity in the general case.
Let $x_0in mathbbR$, we have to show that $lim_xto x_0a^x=a^x_0$. Since $x-x_0to 0$ as $xto x_0$, then $a^x-x_0to1$ by ($1$) and thus
$$lim_xto x_0a^x=lim_xto x_0a^x_0a^x-x_0=a^x_0lim_xto x_0a^x-x_0=a^x_0$$
as was to be shown.
Since $ein mathbbR^>0$, thus $e^x$ is continuous.
answered Mar 23 '14 at 5:38
Jose Antonio
4,36921326
add a comment |Â
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Let $a$ be a positive real number. Then the function $f: mathbbRto mathbbR$ defined by $xmapsto a^x$ is continuous.
Proof:
1) First we prove the continuous at $0$.
We may assume that $a>1$. Let $varepsilon>0$ be arbitrary. Since $a^1/kto 1$ and $a^-1/kto 1$ as $kto infty$, we choose $K$ such that both are $varepsilon$-close to $1$. Let $delta=1/K$, so for $|x|<1/K$ we have
beginalign-1/K<x<1/K\a^-1/K<a^x<a^1/K endalign
which proves that $a^x$ is $varepsilon$-close to $1$ as desired (the case $ale1$ is handled similarly just with the inequality in the other direction).
2) To conclude we prove the continuity in the general case.
Let $x_0in mathbbR$, we have to show that $lim_xto x_0a^x=a^x_0$. Since $x-x_0to 0$ as $xto x_0$, then $a^x-x_0to1$ by ($1$) and thus
$$lim_xto x_0a^x=lim_xto x_0a^x_0a^x-x_0=a^x_0lim_xto x_0a^x-x_0=a^x_0$$
as was to be shown.
Since $ein mathbbR^>0$, thus $e^x$ is continuous.
answered Mar 23 '14 at 5:38
Jose Antonio
4,36921326
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $a$ be a positive real number. Then the function $f: mathbbRto mathbbR$ defined by $xmapsto a^x$ is continuous.
Proof:
1) First we prove the continuous at $0$.
We may assume that $a>1$. Let $varepsilon>0$ be arbitrary. Since $a^1/kto 1$ and $a^-1/kto 1$ as $kto infty$, we choose $K$ such that both are $varepsilon$-close to $1$. Let $delta=1/K$, so for $|x|<1/K$ we have
beginalign-1/K<x<1/K\a^-1/K<a^x<a^1/K endalign
which proves that $a^x$ is $varepsilon$-close to $1$ as desired (the case $ale1$ is handled similarly just with the inequality in the other direction).
2) To conclude we prove the continuity in the general case.
Let $x_0in mathbbR$, we have to show that $lim_xto x_0a^x=a^x_0$. Since $x-x_0to 0$ as $xto x_0$, then $a^x-x_0to1$ by ($1$) and thus
$$lim_xto x_0a^x=lim_xto x_0a^x_0a^x-x_0=a^x_0lim_xto x_0a^x-x_0=a^x_0$$
as was to be shown.
Since $ein mathbbR^>0$, thus $e^x$ is continuous.
answered Mar 23 '14 at 5:38
Jose Antonio
4,36921326
Let $a$ be a positive real number. Then the function $f: mathbbRto mathbbR$ defined by $xmapsto a^x$ is continuous.
Proof:
1) First we prove the continuous at $0$.
We may assume that $a>1$. Let $varepsilon>0$ be arbitrary. Since $a^1/kto 1$ and $a^-1/kto 1$ as $kto infty$, we choose $K$ such that both are $varepsilon$-close to $1$. Let $delta=1/K$, so for $|x|<1/K$ we have
beginalign-1/K<x<1/K\a^-1/K<a^x<a^1/K endalign
which proves that $a^x$ is $varepsilon$-close to $1$ as desired (the case $ale1$ is handled similarly just with the inequality in the other direction).
2) To conclude we prove the continuity in the general case.
Let $x_0in mathbbR$, we have to show that $lim_xto x_0a^x=a^x_0$. Since $x-x_0to 0$ as $xto x_0$, then $a^x-x_0to1$ by ($1$) and thus
$$lim_xto x_0a^x=lim_xto x_0a^x_0a^x-x_0=a^x_0lim_xto x_0a^x-x_0=a^x_0$$
as was to be shown.
Since $ein mathbbR^>0$, thus $e^x$ is continuous.
answered Mar 23 '14 at 5:38
Jose Antonio
4,36921326
answered Mar 23 '14 at 5:38
Jose Antonio
4,36921326
answered Mar 23 '14 at 5:38
Jose Antonio
4,36921326
answered Mar 23 '14 at 5:38
Jose Antonio
4,36921326
4,36921326
add a comment |Â
add a comment |Â
up vote
2
down vote
Start with $|x-c|<delta$.
Let's take a look at the case that $xge c$.
Then:
$$x-c <delta$$
$$x < c+ delta$$
$$e^x < e^c+delta$$
$$0 le e^x-e^c < e^c(e^delta - 1)$$
Let's pick $delta$ such that $varepsilon = e^c(e^delta - 1)$.
Then:
$$delta = ln(1+varepsilon e^-c)$$
Repeat for $x<c$.
In other words, you are able to find a $delta$ for any $varepsilon>0$.
Qed.
answered Mar 23 '14 at 0:47
I like Serena
3,0731718
add a comment |Â
up vote
2
down vote
Start with $|x-c|<delta$.
Let's take a look at the case that $xge c$.
Then:
$$x-c <delta$$
$$x < c+ delta$$
$$e^x < e^c+delta$$
$$0 le e^x-e^c < e^c(e^delta - 1)$$
Let's pick $delta$ such that $varepsilon = e^c(e^delta - 1)$.
Then:
$$delta = ln(1+varepsilon e^-c)$$
Repeat for $x<c$.
In other words, you are able to find a $delta$ for any $varepsilon>0$.
Qed.
answered Mar 23 '14 at 0:47
I like Serena
3,0731718
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Start with $|x-c|<delta$.
Let's take a look at the case that $xge c$.
Then:
$$x-c <delta$$
$$x < c+ delta$$
$$e^x < e^c+delta$$
$$0 le e^x-e^c < e^c(e^delta - 1)$$
Let's pick $delta$ such that $varepsilon = e^c(e^delta - 1)$.
Then:
$$delta = ln(1+varepsilon e^-c)$$
Repeat for $x<c$.
In other words, you are able to find a $delta$ for any $varepsilon>0$.
Qed.
answered Mar 23 '14 at 0:47
I like Serena
3,0731718
Start with $|x-c|<delta$.
Let's take a look at the case that $xge c$.
Then:
$$x-c <delta$$
$$x < c+ delta$$
$$e^x < e^c+delta$$
$$0 le e^x-e^c < e^c(e^delta - 1)$$
Let's pick $delta$ such that $varepsilon = e^c(e^delta - 1)$.
Then:
$$delta = ln(1+varepsilon e^-c)$$
Repeat for $x<c$.
In other words, you are able to find a $delta$ for any $varepsilon>0$.
Qed.
answered Mar 23 '14 at 0:47
I like Serena
3,0731718
answered Mar 23 '14 at 0:47
I like Serena
3,0731718
answered Mar 23 '14 at 0:47
I like Serena
3,0731718
answered Mar 23 '14 at 0:47
I like Serena
3,0731718
3,0731718
add a comment |Â
add a comment |Â
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1
What is known about $e^x$, and how is it defined?
– Daniel Fischer♦
Mar 23 '14 at 0:33
I can see that is is continuous geometrically of course, but not sure how to get this formally
– kiwifruit
Mar 23 '14 at 0:34
1
How to prove it formally depends on what you already know, and which definition you have. The power series definition, $$e^x = sum_n=0^infty fracx^nn!,?$$ Do you have the addition theorem, $e^x+y = e^xcdot e^y$?
– Daniel Fischer♦
Mar 23 '14 at 0:36
1
Apply the definition of continuity everywhere: for all $c$ and for all $epsilon > 0$, there exists $delta > 0$ such that if $|x - c| < delta$, then $|f(x) - f(c)| < epsilon$. So since $e^x$ is defined everywhere on $mathbbR$, let $c in mathbbR$ and let $epsilon > 0$. Can you produce a corresponding $delta > 0$?
– ml0105
Mar 23 '14 at 0:38
2
$lvert e^x-e^crvert = e^ccdotlvert e^x-c-1rvert$. That reduces the problem to showing continuity in $0$.
– Daniel Fischer♦
Mar 23 '14 at 0:42