Prove using limit definition ($varepsilon - delta$)
Let $f:mathbb{R}tomathbb{R}$ be such that $lim_{xto p}$ $f(x) = L$. For convenience, assume $L>0$. Prove, using the $varepsilon-delta$ formalism, that $$lim_{xto p} [f(x)]^2 = L^2$$
My work:
$$lim_{xrightarrow p} [f(x)]^2 = lim_{xto p}f(x) cdotlim_{xto p} f(x) = L cdot L = L^2$$
But that seems too easy, and doesn't involve $varepsilon$ or $delta$. Any other approaches? Please and thank you.
real-analysis limits epsilon-delta
edited Dec 2 at 0:52
mrtaurho
3,40121032
asked Dec 2 at 0:46
Ty Johnson
283
add a comment |
Let $f:mathbb{R}tomathbb{R}$ be such that $lim_{xto p}$ $f(x) = L$. For convenience, assume $L>0$. Prove, using the $varepsilon-delta$ formalism, that $$lim_{xto p} [f(x)]^2 = L^2$$
My work:
$$lim_{xrightarrow p} [f(x)]^2 = lim_{xto p}f(x) cdotlim_{xto p} f(x) = L cdot L = L^2$$
But that seems too easy, and doesn't involve $varepsilon$ or $delta$. Any other approaches? Please and thank you.
real-analysis limits epsilon-delta
edited Dec 2 at 0:52
mrtaurho
3,40121032
asked Dec 2 at 0:46
Ty Johnson
283
2
How do you prove that $lim_{x to p} f(x)g(x) = (lim_{x to p} f(x))(lim_{x to p} g(x))$? (When both limits exist.)
– Trevor Gunn
Dec 2 at 0:51
add a comment |
Let $f:mathbb{R}tomathbb{R}$ be such that $lim_{xto p}$ $f(x) = L$. For convenience, assume $L>0$. Prove, using the $varepsilon-delta$ formalism, that $$lim_{xto p} [f(x)]^2 = L^2$$
My work:
$$lim_{xrightarrow p} [f(x)]^2 = lim_{xto p}f(x) cdotlim_{xto p} f(x) = L cdot L = L^2$$
But that seems too easy, and doesn't involve $varepsilon$ or $delta$. Any other approaches? Please and thank you.
real-analysis limits epsilon-delta
edited Dec 2 at 0:52
mrtaurho
3,40121032
asked Dec 2 at 0:46
Ty Johnson
283
Let $f:mathbb{R}tomathbb{R}$ be such that $lim_{xto p}$ $f(x) = L$. For convenience, assume $L>0$. Prove, using the $varepsilon-delta$ formalism, that $$lim_{xto p} [f(x)]^2 = L^2$$
My work:
$$lim_{xrightarrow p} [f(x)]^2 = lim_{xto p}f(x) cdotlim_{xto p} f(x) = L cdot L = L^2$$
But that seems too easy, and doesn't involve $varepsilon$ or $delta$. Any other approaches? Please and thank you.
real-analysis limits epsilon-delta
real-analysis limits epsilon-delta
edited Dec 2 at 0:52
mrtaurho
3,40121032
asked Dec 2 at 0:46
Ty Johnson
283
edited Dec 2 at 0:52
mrtaurho
3,40121032
asked Dec 2 at 0:46
Ty Johnson
283
edited Dec 2 at 0:52
mrtaurho
3,40121032
edited Dec 2 at 0:52
mrtaurho
3,40121032
edited Dec 2 at 0:52
mrtaurho
3,40121032
3,40121032
asked Dec 2 at 0:46
Ty Johnson
283
asked Dec 2 at 0:46
Ty Johnson
283
asked Dec 2 at 0:46
Ty Johnson
283
283
2
How do you prove that $lim_{x to p} f(x)g(x) = (lim_{x to p} f(x))(lim_{x to p} g(x))$? (When both limits exist.)
– Trevor Gunn
Dec 2 at 0:51
add a comment |
2
How do you prove that $lim_{x to p} f(x)g(x) = (lim_{x to p} f(x))(lim_{x to p} g(x))$? (When both limits exist.)
– Trevor Gunn
Dec 2 at 0:51
2
2
How do you prove that $lim_{x to p} f(x)g(x) = (lim_{x to p} f(x))(lim_{x to p} g(x))$? (When both limits exist.)
– Trevor Gunn
Dec 2 at 0:51
How do you prove that $lim_{x to p} f(x)g(x) = (lim_{x to p} f(x))(lim_{x to p} g(x))$? (When both limits exist.)
– Trevor Gunn
Dec 2 at 0:51
add a comment |
1 Answer
1
active
oldest
votes
Perhaps a more appropriate phrasing for this question would be:
Let $f:mathbb{R}rightarrowmathbb{R}$ be such that $lim_{xrightarrow p}f(x)=L$. For convenience, assume $L>0$. Prove, using the $epsilon-delta$ definition of a limit, that
$$
lim_{xrightarrow p}[f(x)]^2 = L^2
$$
We aren't just looking for some $epsilon$'s and $delta$'s to creep into your work any which way, we want you to unwrap the hypothesis and the desired result with the definition of a limit. So what is that definition?
$lim_{xrightarrow p}f(x)=L$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)-L|<epsilon$.
We haven't proven it yet, but what would our result look like under this definition?
$lim_{xrightarrow p}[f(x)]^2=L^2$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)^2-L^2|<epsilon$.
Please note that the $epsilon$'s and $delta$'s are different in these two definitions!
The game is now to assume the first (since it is our hypothesis) and manipulate it until we get the second. If you need more of a hint, please comment and I will provide some more crumbs.
answered Dec 2 at 1:05
FranklinBash
1012
From limit definition, $|x-p|<delta implies |f(x)-L|<$ ${epsilon}over{|f(x)+L|}$. Now $|f(x)^2-L^2|=|(f(x)-L)(f(x)+L)|<|f(x)-L|cdot|f(x)+L|<$ $epsilonover{|f(x)+L|}$ $cdot|f(x)+L| = epsilon$ Is this it?
– Ty Johnson
Dec 2 at 21:51
Please see above, I need more of a hint, any crumbs will help.
– Ty Johnson
Dec 2 at 21:57
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Perhaps a more appropriate phrasing for this question would be:
Let $f:mathbb{R}rightarrowmathbb{R}$ be such that $lim_{xrightarrow p}f(x)=L$. For convenience, assume $L>0$. Prove, using the $epsilon-delta$ definition of a limit, that
$$
lim_{xrightarrow p}[f(x)]^2 = L^2
$$
We aren't just looking for some $epsilon$'s and $delta$'s to creep into your work any which way, we want you to unwrap the hypothesis and the desired result with the definition of a limit. So what is that definition?
$lim_{xrightarrow p}f(x)=L$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)-L|<epsilon$.
We haven't proven it yet, but what would our result look like under this definition?
$lim_{xrightarrow p}[f(x)]^2=L^2$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)^2-L^2|<epsilon$.
Please note that the $epsilon$'s and $delta$'s are different in these two definitions!
The game is now to assume the first (since it is our hypothesis) and manipulate it until we get the second. If you need more of a hint, please comment and I will provide some more crumbs.
answered Dec 2 at 1:05
FranklinBash
1012
From limit definition, $|x-p|<delta implies |f(x)-L|<$ ${epsilon}over{|f(x)+L|}$. Now $|f(x)^2-L^2|=|(f(x)-L)(f(x)+L)|<|f(x)-L|cdot|f(x)+L|<$ $epsilonover{|f(x)+L|}$ $cdot|f(x)+L| = epsilon$ Is this it?
– Ty Johnson
Dec 2 at 21:51
Please see above, I need more of a hint, any crumbs will help.
– Ty Johnson
Dec 2 at 21:57
add a comment |
Perhaps a more appropriate phrasing for this question would be:
Let $f:mathbb{R}rightarrowmathbb{R}$ be such that $lim_{xrightarrow p}f(x)=L$. For convenience, assume $L>0$. Prove, using the $epsilon-delta$ definition of a limit, that
$$
lim_{xrightarrow p}[f(x)]^2 = L^2
$$
We aren't just looking for some $epsilon$'s and $delta$'s to creep into your work any which way, we want you to unwrap the hypothesis and the desired result with the definition of a limit. So what is that definition?
$lim_{xrightarrow p}f(x)=L$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)-L|<epsilon$.
We haven't proven it yet, but what would our result look like under this definition?
$lim_{xrightarrow p}[f(x)]^2=L^2$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)^2-L^2|<epsilon$.
Please note that the $epsilon$'s and $delta$'s are different in these two definitions!
The game is now to assume the first (since it is our hypothesis) and manipulate it until we get the second. If you need more of a hint, please comment and I will provide some more crumbs.
answered Dec 2 at 1:05
FranklinBash
1012
From limit definition, $|x-p|<delta implies |f(x)-L|<$ ${epsilon}over{|f(x)+L|}$. Now $|f(x)^2-L^2|=|(f(x)-L)(f(x)+L)|<|f(x)-L|cdot|f(x)+L|<$ $epsilonover{|f(x)+L|}$ $cdot|f(x)+L| = epsilon$ Is this it?
– Ty Johnson
Dec 2 at 21:51
Please see above, I need more of a hint, any crumbs will help.
– Ty Johnson
Dec 2 at 21:57
add a comment |
Perhaps a more appropriate phrasing for this question would be:
Let $f:mathbb{R}rightarrowmathbb{R}$ be such that $lim_{xrightarrow p}f(x)=L$. For convenience, assume $L>0$. Prove, using the $epsilon-delta$ definition of a limit, that
$$
lim_{xrightarrow p}[f(x)]^2 = L^2
$$
We aren't just looking for some $epsilon$'s and $delta$'s to creep into your work any which way, we want you to unwrap the hypothesis and the desired result with the definition of a limit. So what is that definition?
$lim_{xrightarrow p}f(x)=L$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)-L|<epsilon$.
We haven't proven it yet, but what would our result look like under this definition?
$lim_{xrightarrow p}[f(x)]^2=L^2$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)^2-L^2|<epsilon$.
Please note that the $epsilon$'s and $delta$'s are different in these two definitions!
The game is now to assume the first (since it is our hypothesis) and manipulate it until we get the second. If you need more of a hint, please comment and I will provide some more crumbs.
answered Dec 2 at 1:05
FranklinBash
1012
Perhaps a more appropriate phrasing for this question would be:
Let $f:mathbb{R}rightarrowmathbb{R}$ be such that $lim_{xrightarrow p}f(x)=L$. For convenience, assume $L>0$. Prove, using the $epsilon-delta$ definition of a limit, that
$$
lim_{xrightarrow p}[f(x)]^2 = L^2
$$
We aren't just looking for some $epsilon$'s and $delta$'s to creep into your work any which way, we want you to unwrap the hypothesis and the desired result with the definition of a limit. So what is that definition?
$lim_{xrightarrow p}f(x)=L$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)-L|<epsilon$.
We haven't proven it yet, but what would our result look like under this definition?
$lim_{xrightarrow p}[f(x)]^2=L^2$ if and only if $forallepsilon>0$ $existsdelta>0$ such that if $0<|x-p|<delta$ then $|f(x)^2-L^2|<epsilon$.
Please note that the $epsilon$'s and $delta$'s are different in these two definitions!
The game is now to assume the first (since it is our hypothesis) and manipulate it until we get the second. If you need more of a hint, please comment and I will provide some more crumbs.
answered Dec 2 at 1:05
FranklinBash
1012
answered Dec 2 at 1:05
FranklinBash
1012
answered Dec 2 at 1:05
FranklinBash
1012
answered Dec 2 at 1:05
FranklinBash
1012
1012
From limit definition, $|x-p|<delta implies |f(x)-L|<$ ${epsilon}over{|f(x)+L|}$. Now $|f(x)^2-L^2|=|(f(x)-L)(f(x)+L)|<|f(x)-L|cdot|f(x)+L|<$ $epsilonover{|f(x)+L|}$ $cdot|f(x)+L| = epsilon$ Is this it?
– Ty Johnson
Dec 2 at 21:51
Please see above, I need more of a hint, any crumbs will help.
– Ty Johnson
Dec 2 at 21:57
add a comment |
From limit definition, $|x-p|<delta implies |f(x)-L|<$ ${epsilon}over{|f(x)+L|}$. Now $|f(x)^2-L^2|=|(f(x)-L)(f(x)+L)|<|f(x)-L|cdot|f(x)+L|<$ $epsilonover{|f(x)+L|}$ $cdot|f(x)+L| = epsilon$ Is this it?
– Ty Johnson
Dec 2 at 21:51
Please see above, I need more of a hint, any crumbs will help.
– Ty Johnson
Dec 2 at 21:57
From limit definition, $|x-p|<delta implies |f(x)-L|<$ ${epsilon}over{|f(x)+L|}$. Now $|f(x)^2-L^2|=|(f(x)-L)(f(x)+L)|<|f(x)-L|cdot|f(x)+L|<$ $epsilonover{|f(x)+L|}$ $cdot|f(x)+L| = epsilon$ Is this it?
– Ty Johnson
Dec 2 at 21:51
From limit definition, $|x-p|<delta implies |f(x)-L|<$ ${epsilon}over{|f(x)+L|}$. Now $|f(x)^2-L^2|=|(f(x)-L)(f(x)+L)|<|f(x)-L|cdot|f(x)+L|<$ $epsilonover{|f(x)+L|}$ $cdot|f(x)+L| = epsilon$ Is this it?
– Ty Johnson
Dec 2 at 21:51
Please see above, I need more of a hint, any crumbs will help.
– Ty Johnson
Dec 2 at 21:57
Please see above, I need more of a hint, any crumbs will help.
– Ty Johnson
Dec 2 at 21:57
add a comment |
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2
How do you prove that $lim_{x to p} f(x)g(x) = (lim_{x to p} f(x))(lim_{x to p} g(x))$? (When both limits exist.)
– Trevor Gunn
Dec 2 at 0:51