Mixing
Use definition of limit to prove $lim_ {t to a} f(t^2) =L$
$begingroup$
Suppose:
$$lim_ {x to a^2} f(x) =L $$
Use this to prove $lim_{t to a} f(t^2) =L$.
This is evident from the limit substitution rule, but in this course we have not covered this and we should use the formal definition of the limit. That is:
$$ forall epsilon>0 quad exists delta quad s.t. quad |x-a^2|< delta implies |f(x)-L|< epsilon$$
must be used to prove:
$$ forall epsilon>0 quad exists delta quad s.t. quad |t-a|< delta implies |f(t^2)-L|< epsilon$$
how do I make this connection.
real-analysis limits epsilon-delta
asked Jan 28 at 9:44
Wesley StrikWesley Strik
2,194424
$endgroup$
add a comment |
$begingroup$
Suppose:
$$lim_ {x to a^2} f(x) =L $$
Use this to prove $lim_{t to a} f(t^2) =L$.
This is evident from the limit substitution rule, but in this course we have not covered this and we should use the formal definition of the limit. That is:
$$ forall epsilon>0 quad exists delta quad s.t. quad |x-a^2|< delta implies |f(x)-L|< epsilon$$
must be used to prove:
$$ forall epsilon>0 quad exists delta quad s.t. quad |t-a|< delta implies |f(t^2)-L|< epsilon$$
how do I make this connection.
real-analysis limits epsilon-delta
asked Jan 28 at 9:44
Wesley StrikWesley Strik
2,194424
$endgroup$
1
$begingroup$
Maybe this can follow from a more general result which is not so hard to prove… (substitution theorem for limits)
$endgroup$
– James
Jan 28 at 9:46
$begingroup$
What is that $displaystylelim_{ntoinfty}$ doing in the title?
$endgroup$
– José Carlos Santos
Jan 28 at 9:46
add a comment |
$begingroup$
Suppose:
$$lim_ {x to a^2} f(x) =L $$
Use this to prove $lim_{t to a} f(t^2) =L$.
This is evident from the limit substitution rule, but in this course we have not covered this and we should use the formal definition of the limit. That is:
$$ forall epsilon>0 quad exists delta quad s.t. quad |x-a^2|< delta implies |f(x)-L|< epsilon$$
must be used to prove:
$$ forall epsilon>0 quad exists delta quad s.t. quad |t-a|< delta implies |f(t^2)-L|< epsilon$$
how do I make this connection.
real-analysis limits epsilon-delta
asked Jan 28 at 9:44
Wesley StrikWesley Strik
2,194424
$endgroup$
Suppose:
$$lim_ {x to a^2} f(x) =L $$
Use this to prove $lim_{t to a} f(t^2) =L$.
This is evident from the limit substitution rule, but in this course we have not covered this and we should use the formal definition of the limit. That is:
$$ forall epsilon>0 quad exists delta quad s.t. quad |x-a^2|< delta implies |f(x)-L|< epsilon$$
must be used to prove:
$$ forall epsilon>0 quad exists delta quad s.t. quad |t-a|< delta implies |f(t^2)-L|< epsilon$$
how do I make this connection.
real-analysis limits epsilon-delta
real-analysis limits epsilon-delta
asked Jan 28 at 9:44
Wesley StrikWesley Strik
2,194424
asked Jan 28 at 9:44
Wesley StrikWesley Strik
2,194424
edited Jan 28 at 9:50
Wesley Strik
asked Jan 28 at 9:44
Wesley StrikWesley Strik
2,194424
asked Jan 28 at 9:44
Wesley StrikWesley Strik
2,194424
asked Jan 28 at 9:44
Wesley StrikWesley Strik
2,194424
2,194424
1
$begingroup$
Maybe this can follow from a more general result which is not so hard to prove… (substitution theorem for limits)
$endgroup$
– James
Jan 28 at 9:46
$begingroup$
What is that $displaystylelim_{ntoinfty}$ doing in the title?
$endgroup$
– José Carlos Santos
Jan 28 at 9:46
add a comment |
1
$begingroup$
Maybe this can follow from a more general result which is not so hard to prove… (substitution theorem for limits)
$endgroup$
– James
Jan 28 at 9:46
$begingroup$
What is that $displaystylelim_{ntoinfty}$ doing in the title?
$endgroup$
– José Carlos Santos
Jan 28 at 9:46
1
1
$begingroup$
Maybe this can follow from a more general result which is not so hard to prove… (substitution theorem for limits)
$endgroup$
– James
Jan 28 at 9:46
$begingroup$
Maybe this can follow from a more general result which is not so hard to prove… (substitution theorem for limits)
$endgroup$
– James
Jan 28 at 9:46
$begingroup$
What is that $displaystylelim_{ntoinfty}$ doing in the title?
$endgroup$
– José Carlos Santos
Jan 28 at 9:46
$begingroup$
What is that $displaystylelim_{ntoinfty}$ doing in the title?
$endgroup$
– José Carlos Santos
Jan 28 at 9:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$|t-a| <delta'$ implies $|t^{2}-a^{2}|=|t-a||t+a| <delta' (|t|+|a|)<delta' (delta'+2|a|)$. So choose $delta'$ such that $delta' (delta'+2|a|) <delta$. It is enough to take $delta' <1$ and $delta' <frac {delta} {1+2|a|}$. Then $|t-a| <delta'$ implies $|f(t^{2})-L| <epsilon$.
answered Jan 28 at 9:49
Kavi Rama MurthyKavi Rama Murthy
70.6k53170
$endgroup$
$begingroup$
note to self: set $x=t^2$ and it follows.
$endgroup$
– Wesley Strik
Jan 28 at 10:04
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
$|t-a| <delta'$ implies $|t^{2}-a^{2}|=|t-a||t+a| <delta' (|t|+|a|)<delta' (delta'+2|a|)$. So choose $delta'$ such that $delta' (delta'+2|a|) <delta$. It is enough to take $delta' <1$ and $delta' <frac {delta} {1+2|a|}$. Then $|t-a| <delta'$ implies $|f(t^{2})-L| <epsilon$.
answered Jan 28 at 9:49
Kavi Rama MurthyKavi Rama Murthy
70.6k53170
$endgroup$
$begingroup$
note to self: set $x=t^2$ and it follows.
$endgroup$
– Wesley Strik
Jan 28 at 10:04
add a comment |
$begingroup$
$|t-a| <delta'$ implies $|t^{2}-a^{2}|=|t-a||t+a| <delta' (|t|+|a|)<delta' (delta'+2|a|)$. So choose $delta'$ such that $delta' (delta'+2|a|) <delta$. It is enough to take $delta' <1$ and $delta' <frac {delta} {1+2|a|}$. Then $|t-a| <delta'$ implies $|f(t^{2})-L| <epsilon$.
answered Jan 28 at 9:49
Kavi Rama MurthyKavi Rama Murthy
70.6k53170
$endgroup$
$begingroup$
note to self: set $x=t^2$ and it follows.
$endgroup$
– Wesley Strik
Jan 28 at 10:04
add a comment |
$begingroup$
$|t-a| <delta'$ implies $|t^{2}-a^{2}|=|t-a||t+a| <delta' (|t|+|a|)<delta' (delta'+2|a|)$. So choose $delta'$ such that $delta' (delta'+2|a|) <delta$. It is enough to take $delta' <1$ and $delta' <frac {delta} {1+2|a|}$. Then $|t-a| <delta'$ implies $|f(t^{2})-L| <epsilon$.
answered Jan 28 at 9:49
Kavi Rama MurthyKavi Rama Murthy
70.6k53170
$endgroup$
$|t-a| <delta'$ implies $|t^{2}-a^{2}|=|t-a||t+a| <delta' (|t|+|a|)<delta' (delta'+2|a|)$. So choose $delta'$ such that $delta' (delta'+2|a|) <delta$. It is enough to take $delta' <1$ and $delta' <frac {delta} {1+2|a|}$. Then $|t-a| <delta'$ implies $|f(t^{2})-L| <epsilon$.
answered Jan 28 at 9:49
Kavi Rama MurthyKavi Rama Murthy
70.6k53170
answered Jan 28 at 9:49
Kavi Rama MurthyKavi Rama Murthy
70.6k53170
answered Jan 28 at 9:49
Kavi Rama MurthyKavi Rama Murthy
70.6k53170
answered Jan 28 at 9:49
Kavi Rama MurthyKavi Rama Murthy
70.6k53170
70.6k53170
$begingroup$
note to self: set $x=t^2$ and it follows.
$endgroup$
– Wesley Strik
Jan 28 at 10:04
add a comment |
$begingroup$
note to self: set $x=t^2$ and it follows.
$endgroup$
– Wesley Strik
Jan 28 at 10:04
$begingroup$
note to self: set $x=t^2$ and it follows.
$endgroup$
– Wesley Strik
Jan 28 at 10:04
$begingroup$
note to self: set $x=t^2$ and it follows.
$endgroup$
– Wesley Strik
Jan 28 at 10:04
add a comment |
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1
$begingroup$
Maybe this can follow from a more general result which is not so hard to prove… (substitution theorem for limits)
$endgroup$
– James
Jan 28 at 9:46
$begingroup$
What is that $displaystylelim_{ntoinfty}$ doing in the title?
$endgroup$
– José Carlos Santos
Jan 28 at 9:46