Prove using limit definition ($varepsilon - delta$)












0












$begingroup$


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







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  • 2




    $begingroup$
    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.)
    $endgroup$
    – Trevor Gunn
    Dec 2 '18 at 0:51


















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    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.)
    $endgroup$
    – Trevor Gunn
    Dec 2 '18 at 0:51
















0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






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edited Dec 2 '18 at 0:52









mrtaurho

4,05921234




4,05921234










asked Dec 2 '18 at 0:46









Ty JohnsonTy Johnson

323




323








  • 2




    $begingroup$
    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.)
    $endgroup$
    – Trevor Gunn
    Dec 2 '18 at 0:51
















  • 2




    $begingroup$
    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.)
    $endgroup$
    – Trevor Gunn
    Dec 2 '18 at 0:51










2




2




$begingroup$
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.)
$endgroup$
– Trevor Gunn
Dec 2 '18 at 0:51






$begingroup$
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.)
$endgroup$
– Trevor Gunn
Dec 2 '18 at 0:51












1 Answer
1






active

oldest

votes


















4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – Ty Johnson
    Dec 2 '18 at 21:51












  • $begingroup$
    Please see above, I need more of a hint, any crumbs will help.
    $endgroup$
    – Ty Johnson
    Dec 2 '18 at 21:57











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4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – Ty Johnson
    Dec 2 '18 at 21:51












  • $begingroup$
    Please see above, I need more of a hint, any crumbs will help.
    $endgroup$
    – Ty Johnson
    Dec 2 '18 at 21:57
















4












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – Ty Johnson
    Dec 2 '18 at 21:51












  • $begingroup$
    Please see above, I need more of a hint, any crumbs will help.
    $endgroup$
    – Ty Johnson
    Dec 2 '18 at 21:57














4












4








4





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 '18 at 1:05









FranklinBashFranklinBash

1012




1012












  • $begingroup$
    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?
    $endgroup$
    – Ty Johnson
    Dec 2 '18 at 21:51












  • $begingroup$
    Please see above, I need more of a hint, any crumbs will help.
    $endgroup$
    – Ty Johnson
    Dec 2 '18 at 21:57


















  • $begingroup$
    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?
    $endgroup$
    – Ty Johnson
    Dec 2 '18 at 21:51












  • $begingroup$
    Please see above, I need more of a hint, any crumbs will help.
    $endgroup$
    – Ty Johnson
    Dec 2 '18 at 21:57
















$begingroup$
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?
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:51






$begingroup$
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?
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:51














$begingroup$
Please see above, I need more of a hint, any crumbs will help.
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:57




$begingroup$
Please see above, I need more of a hint, any crumbs will help.
$endgroup$
– Ty Johnson
Dec 2 '18 at 21:57


















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