Mixing
Is this a Viable Strategy to Prove any Polynomial Limit?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Find $Q(x)=|f(x)-L|div|x-c|$
Assume $|x-c|<delta$, so that
$$|x-c| cdot Q(x)<delta cdot Q(x)$$$Q(x)$ is of the form, $ax^n+bx^n-1+dx^n-2+ ...$, so, applying the Triangle Inequality,
$$delta cdot Q(x) = delta cdot (ax^n+bx^n-1+dx^n-2+ ...) le delta cdot (|a||x|^n+|b||x|^n-1+|d||x|^n-2+ ...)$$- Assume $|x-c|<1$, so that
$$-1<x-c<1 implies c-1<x<c+1 implies |x| < M = maxc-1$$ - Then,
$$delta cdot (a|x|^n+b|x|^n-1+d|x|^n-2+ ...) lt delta cdot (aM^n+bM^n-1+dM^n-2+ ...) = delta cdot K$$ - Set $delta=max1, epsilon/K$.
Have I made any glaring mistakes here?
limits proof-verification epsilon-delta
asked Aug 5 at 22:57
Vpie649
915
add a comment |Â
up vote
0
down vote
favorite
Find $Q(x)=|f(x)-L|div|x-c|$
Assume $|x-c|<delta$, so that
$$|x-c| cdot Q(x)<delta cdot Q(x)$$$Q(x)$ is of the form, $ax^n+bx^n-1+dx^n-2+ ...$, so, applying the Triangle Inequality,
$$delta cdot Q(x) = delta cdot (ax^n+bx^n-1+dx^n-2+ ...) le delta cdot (|a||x|^n+|b||x|^n-1+|d||x|^n-2+ ...)$$- Assume $|x-c|<1$, so that
$$-1<x-c<1 implies c-1<x<c+1 implies |x| < M = maxc-1$$ - Then,
$$delta cdot (a|x|^n+b|x|^n-1+d|x|^n-2+ ...) lt delta cdot (aM^n+bM^n-1+dM^n-2+ ...) = delta cdot K$$ - Set $delta=max1, epsilon/K$.
Have I made any glaring mistakes here?
limits proof-verification epsilon-delta
asked Aug 5 at 22:57
Vpie649
915
What are you trying to prove?
– Kavi Rama Murthy
Aug 5 at 23:30
@KaviRamaMurthy $lim_x to cf(x) = L$
– Vpie649
Aug 5 at 23:37
1
You have not made any mistake.
– Kavi Rama Murthy
Aug 5 at 23:40
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find $Q(x)=|f(x)-L|div|x-c|$
Assume $|x-c|<delta$, so that
$$|x-c| cdot Q(x)<delta cdot Q(x)$$$Q(x)$ is of the form, $ax^n+bx^n-1+dx^n-2+ ...$, so, applying the Triangle Inequality,
$$delta cdot Q(x) = delta cdot (ax^n+bx^n-1+dx^n-2+ ...) le delta cdot (|a||x|^n+|b||x|^n-1+|d||x|^n-2+ ...)$$- Assume $|x-c|<1$, so that
$$-1<x-c<1 implies c-1<x<c+1 implies |x| < M = maxc-1$$ - Then,
$$delta cdot (a|x|^n+b|x|^n-1+d|x|^n-2+ ...) lt delta cdot (aM^n+bM^n-1+dM^n-2+ ...) = delta cdot K$$ - Set $delta=max1, epsilon/K$.
Have I made any glaring mistakes here?
limits proof-verification epsilon-delta
asked Aug 5 at 22:57
Vpie649
915
Find $Q(x)=|f(x)-L|div|x-c|$
Assume $|x-c|<delta$, so that
$$|x-c| cdot Q(x)<delta cdot Q(x)$$$Q(x)$ is of the form, $ax^n+bx^n-1+dx^n-2+ ...$, so, applying the Triangle Inequality,
$$delta cdot Q(x) = delta cdot (ax^n+bx^n-1+dx^n-2+ ...) le delta cdot (|a||x|^n+|b||x|^n-1+|d||x|^n-2+ ...)$$- Assume $|x-c|<1$, so that
$$-1<x-c<1 implies c-1<x<c+1 implies |x| < M = maxc-1$$ - Then,
$$delta cdot (a|x|^n+b|x|^n-1+d|x|^n-2+ ...) lt delta cdot (aM^n+bM^n-1+dM^n-2+ ...) = delta cdot K$$ - Set $delta=max1, epsilon/K$.
Have I made any glaring mistakes here?
limits proof-verification epsilon-delta
asked Aug 5 at 22:57
Vpie649
915
asked Aug 5 at 22:57
Vpie649
915
asked Aug 5 at 22:57
Vpie649
915
asked Aug 5 at 22:57
Vpie649
915
915
What are you trying to prove?
– Kavi Rama Murthy
Aug 5 at 23:30
@KaviRamaMurthy $lim_x to cf(x) = L$
– Vpie649
Aug 5 at 23:37
1
You have not made any mistake.
– Kavi Rama Murthy
Aug 5 at 23:40
add a comment |Â
What are you trying to prove?
– Kavi Rama Murthy
Aug 5 at 23:30
@KaviRamaMurthy $lim_x to cf(x) = L$
– Vpie649
Aug 5 at 23:37
1
You have not made any mistake.
– Kavi Rama Murthy
Aug 5 at 23:40
What are you trying to prove?
– Kavi Rama Murthy
Aug 5 at 23:30
What are you trying to prove?
– Kavi Rama Murthy
Aug 5 at 23:30
@KaviRamaMurthy $lim_x to cf(x) = L$
– Vpie649
Aug 5 at 23:37
@KaviRamaMurthy $lim_x to cf(x) = L$
– Vpie649
Aug 5 at 23:37
1
1
You have not made any mistake.
– Kavi Rama Murthy
Aug 5 at 23:40
You have not made any mistake.
– Kavi Rama Murthy
Aug 5 at 23:40
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873407%2fis-this-a-viable-strategy-to-prove-any-polynomial-limit%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
What are you trying to prove?
– Kavi Rama Murthy
Aug 5 at 23:30
@KaviRamaMurthy $lim_x to cf(x) = L$
– Vpie649
Aug 5 at 23:37
1
You have not made any mistake.
– Kavi Rama Murthy
Aug 5 at 23:40