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Finding $delta $ for $lim_x to 0 x [3-cos(x^2)] = 0 $
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When reading Spivak's calculus book, I stumbled upon this limit: $$ lim_x to 0 x [3-cos(x^2)] = 0 $$
Proof:
We have that $$0 < |x| < delta $$ Also
$$ |x (3-cos(x^2))| = |x| |3-cos(x^2)| < epsilon$$
Since,
$$ |3-cos(x^2)| le 4$$
we can write that:
$$ |x| |3-cos(x^2)| < 4|x| < epsilon$$
$$therefore |x| < epsilon / 4$$
$$therefore delta = epsilon / 4$$
By letting $delta = epsilon / 4$, we get that $|x (3-cos(x^2))|<epsilon$ if $0 < |x| < delta $.
Thus, $ lim_x to 0 x [3-cos(x^2)] = 0 $.
Since I don't have the book with answers, I can't verify my solution (the book is on its way).
Is there anything that I missed here? Maybe there is another way, more concise perhaps, of finding a $delta $?
calculus real-analysis limits
edited Jul 15 at 4:53
Michael Hardy
204k23186463
asked Jul 15 at 4:41
nullbyte
1203
add a comment |Â
up vote
2
down vote
favorite
When reading Spivak's calculus book, I stumbled upon this limit: $$ lim_x to 0 x [3-cos(x^2)] = 0 $$
Proof:
We have that $$0 < |x| < delta $$ Also
$$ |x (3-cos(x^2))| = |x| |3-cos(x^2)| < epsilon$$
Since,
$$ |3-cos(x^2)| le 4$$
we can write that:
$$ |x| |3-cos(x^2)| < 4|x| < epsilon$$
$$therefore |x| < epsilon / 4$$
$$therefore delta = epsilon / 4$$
By letting $delta = epsilon / 4$, we get that $|x (3-cos(x^2))|<epsilon$ if $0 < |x| < delta $.
Thus, $ lim_x to 0 x [3-cos(x^2)] = 0 $.
Since I don't have the book with answers, I can't verify my solution (the book is on its way).
Is there anything that I missed here? Maybe there is another way, more concise perhaps, of finding a $delta $?
calculus real-analysis limits
edited Jul 15 at 4:53
Michael Hardy
204k23186463
asked Jul 15 at 4:41
nullbyte
1203
1
I think it's correct.
– Nosrati
Jul 15 at 4:53
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
When reading Spivak's calculus book, I stumbled upon this limit: $$ lim_x to 0 x [3-cos(x^2)] = 0 $$
Proof:
We have that $$0 < |x| < delta $$ Also
$$ |x (3-cos(x^2))| = |x| |3-cos(x^2)| < epsilon$$
Since,
$$ |3-cos(x^2)| le 4$$
we can write that:
$$ |x| |3-cos(x^2)| < 4|x| < epsilon$$
$$therefore |x| < epsilon / 4$$
$$therefore delta = epsilon / 4$$
By letting $delta = epsilon / 4$, we get that $|x (3-cos(x^2))|<epsilon$ if $0 < |x| < delta $.
Thus, $ lim_x to 0 x [3-cos(x^2)] = 0 $.
Since I don't have the book with answers, I can't verify my solution (the book is on its way).
Is there anything that I missed here? Maybe there is another way, more concise perhaps, of finding a $delta $?
calculus real-analysis limits
edited Jul 15 at 4:53
Michael Hardy
204k23186463
asked Jul 15 at 4:41
nullbyte
1203
When reading Spivak's calculus book, I stumbled upon this limit: $$ lim_x to 0 x [3-cos(x^2)] = 0 $$
Proof:
We have that $$0 < |x| < delta $$ Also
$$ |x (3-cos(x^2))| = |x| |3-cos(x^2)| < epsilon$$
Since,
$$ |3-cos(x^2)| le 4$$
we can write that:
$$ |x| |3-cos(x^2)| < 4|x| < epsilon$$
$$therefore |x| < epsilon / 4$$
$$therefore delta = epsilon / 4$$
By letting $delta = epsilon / 4$, we get that $|x (3-cos(x^2))|<epsilon$ if $0 < |x| < delta $.
Thus, $ lim_x to 0 x [3-cos(x^2)] = 0 $.
Since I don't have the book with answers, I can't verify my solution (the book is on its way).
Is there anything that I missed here? Maybe there is another way, more concise perhaps, of finding a $delta $?
calculus real-analysis limits
edited Jul 15 at 4:53
Michael Hardy
204k23186463
asked Jul 15 at 4:41
nullbyte
1203
edited Jul 15 at 4:53
Michael Hardy
204k23186463
edited Jul 15 at 4:53
Michael Hardy
204k23186463
edited Jul 15 at 4:53
Michael Hardy
204k23186463
204k23186463
asked Jul 15 at 4:41
nullbyte
1203
asked Jul 15 at 4:41
nullbyte
1203
asked Jul 15 at 4:41
nullbyte
1203
1203
1
I think it's correct.
– Nosrati
Jul 15 at 4:53
add a comment |Â
1
I think it's correct.
– Nosrati
Jul 15 at 4:53
1
1
I think it's correct.
– Nosrati
Jul 15 at 4:53
I think it's correct.
– Nosrati
Jul 15 at 4:53
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your answer is absolutely correct.
answered Jul 15 at 4:59
MichaelCarrick
1077
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your answer is absolutely correct.
answered Jul 15 at 4:59
MichaelCarrick
1077
add a comment |Â
up vote
1
down vote
accepted
Your answer is absolutely correct.
answered Jul 15 at 4:59
MichaelCarrick
1077
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your answer is absolutely correct.
answered Jul 15 at 4:59
MichaelCarrick
1077
Your answer is absolutely correct.
answered Jul 15 at 4:59
MichaelCarrick
1077
answered Jul 15 at 4:59
MichaelCarrick
1077
answered Jul 15 at 4:59
MichaelCarrick
1077
answered Jul 15 at 4:59
MichaelCarrick
1077
1077
add a comment |Â
add a comment |Â
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1
I think it's correct.
– Nosrati
Jul 15 at 4:53