Mixing
Proving Limits at Negative Infinity.
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Prove the following using the definition of a limit:
$$lim_{x rightarrow -infty} (1 + x^3) = -infty.$$
I know we have to show that for all $x < N$, there must be $f(x) < M$, but I'm not quite sure what to do from here.
Any help would be greatly appreciated,
Thank you.
limits epsilon-delta
edited Oct 8 '14 at 16:49
Git Gud
28.8k1050100
asked Oct 8 '14 at 16:22
Ian MurphyIan Murphy
10111
$endgroup$
add a comment |
$begingroup$
Prove the following using the definition of a limit:
$$lim_{x rightarrow -infty} (1 + x^3) = -infty.$$
I know we have to show that for all $x < N$, there must be $f(x) < M$, but I'm not quite sure what to do from here.
Any help would be greatly appreciated,
Thank you.
limits epsilon-delta
edited Oct 8 '14 at 16:49
Git Gud
28.8k1050100
asked Oct 8 '14 at 16:22
Ian MurphyIan Murphy
10111
$endgroup$
$begingroup$
Could you, perhaps, show us what you've attempted?
$endgroup$
– Robin Goodfellow
Oct 8 '14 at 16:25
$begingroup$
If $x < -100$, what can you say about $1 + x^3$?
$endgroup$
– Pedro M.
Oct 8 '14 at 16:28
add a comment |
$begingroup$
Prove the following using the definition of a limit:
$$lim_{x rightarrow -infty} (1 + x^3) = -infty.$$
I know we have to show that for all $x < N$, there must be $f(x) < M$, but I'm not quite sure what to do from here.
Any help would be greatly appreciated,
Thank you.
limits epsilon-delta
edited Oct 8 '14 at 16:49
Git Gud
28.8k1050100
asked Oct 8 '14 at 16:22
Ian MurphyIan Murphy
10111
$endgroup$
Prove the following using the definition of a limit:
$$lim_{x rightarrow -infty} (1 + x^3) = -infty.$$
I know we have to show that for all $x < N$, there must be $f(x) < M$, but I'm not quite sure what to do from here.
Any help would be greatly appreciated,
Thank you.
limits epsilon-delta
limits epsilon-delta
edited Oct 8 '14 at 16:49
Git Gud
28.8k1050100
asked Oct 8 '14 at 16:22
Ian MurphyIan Murphy
10111
edited Oct 8 '14 at 16:49
Git Gud
28.8k1050100
asked Oct 8 '14 at 16:22
Ian MurphyIan Murphy
10111
edited Oct 8 '14 at 16:49
Git Gud
28.8k1050100
edited Oct 8 '14 at 16:49
Git Gud
28.8k1050100
edited Oct 8 '14 at 16:49
Git Gud
28.8k1050100
28.8k1050100
asked Oct 8 '14 at 16:22
Ian MurphyIan Murphy
10111
asked Oct 8 '14 at 16:22
Ian MurphyIan Murphy
10111
asked Oct 8 '14 at 16:22
Ian MurphyIan Murphy
10111
10111
$begingroup$
Could you, perhaps, show us what you've attempted?
$endgroup$
– Robin Goodfellow
Oct 8 '14 at 16:25
$begingroup$
If $x < -100$, what can you say about $1 + x^3$?
$endgroup$
– Pedro M.
Oct 8 '14 at 16:28
add a comment |
$begingroup$
Could you, perhaps, show us what you've attempted?
$endgroup$
– Robin Goodfellow
Oct 8 '14 at 16:25
$begingroup$
If $x < -100$, what can you say about $1 + x^3$?
$endgroup$
– Pedro M.
Oct 8 '14 at 16:28
$begingroup$
Could you, perhaps, show us what you've attempted?
$endgroup$
– Robin Goodfellow
Oct 8 '14 at 16:25
$begingroup$
Could you, perhaps, show us what you've attempted?
$endgroup$
– Robin Goodfellow
Oct 8 '14 at 16:25
$begingroup$
If $x < -100$, what can you say about $1 + x^3$?
$endgroup$
– Pedro M.
Oct 8 '14 at 16:28
$begingroup$
If $x < -100$, what can you say about $1 + x^3$?
$endgroup$
– Pedro M.
Oct 8 '14 at 16:28
add a comment |
1 Answer
1
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oldest
votes
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Let $M < 0$ be given, choose N such that $N < sqrt[3]{M-1}$. For $x < N$, we have: $1 + x^3 < 1 + (M - 1) = M$. This proves the result.
answered Oct 8 '14 at 16:29
DeepSeaDeepSea
71.2k54487
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Let $M < 0$ be given, choose N such that $N < sqrt[3]{M-1}$. For $x < N$, we have: $1 + x^3 < 1 + (M - 1) = M$. This proves the result.
answered Oct 8 '14 at 16:29
DeepSeaDeepSea
71.2k54487
$endgroup$
add a comment |
$begingroup$
Let $M < 0$ be given, choose N such that $N < sqrt[3]{M-1}$. For $x < N$, we have: $1 + x^3 < 1 + (M - 1) = M$. This proves the result.
answered Oct 8 '14 at 16:29
DeepSeaDeepSea
71.2k54487
$endgroup$
add a comment |
$begingroup$
Let $M < 0$ be given, choose N such that $N < sqrt[3]{M-1}$. For $x < N$, we have: $1 + x^3 < 1 + (M - 1) = M$. This proves the result.
answered Oct 8 '14 at 16:29
DeepSeaDeepSea
71.2k54487
$endgroup$
Let $M < 0$ be given, choose N such that $N < sqrt[3]{M-1}$. For $x < N$, we have: $1 + x^3 < 1 + (M - 1) = M$. This proves the result.
answered Oct 8 '14 at 16:29
DeepSeaDeepSea
71.2k54487
answered Oct 8 '14 at 16:29
DeepSeaDeepSea
71.2k54487
answered Oct 8 '14 at 16:29
DeepSeaDeepSea
71.2k54487
answered Oct 8 '14 at 16:29
DeepSeaDeepSea
71.2k54487
71.2k54487
add a comment |
add a comment |
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$begingroup$
Could you, perhaps, show us what you've attempted?
$endgroup$
– Robin Goodfellow
Oct 8 '14 at 16:25
$begingroup$
If $x < -100$, what can you say about $1 + x^3$?
$endgroup$
– Pedro M.
Oct 8 '14 at 16:28