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[Stuck]: Final step of solving limit, Calculus I
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Calculate the value of $k$ such that the following limit has a finite solution, $L$ such that $L ne 0$:
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos(x)-1)x^k$$
I use the Taylor Series expansions of $e^x$ and $cos(x)$ and simplify the above expression to the following:
$$lim_xrightarrow0 frac-frac14x^6+(frac148-frac112)x^8+(frac1144)x^10x^k$$
Now I have a mental roadblock. For $k<6$ the above limit goes to zero and for $k>6$ this expression should diverge but in my head it goes to zero again.... I am trying to think of a simple example to convince myself but can't. Can someone please help me understand this?
Thanks.
calculus limits taylor-expansion
edited 22 hours ago
Wong Austin
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asked 22 hours ago
Calum
333
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Calum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
5
down vote
favorite
Calculate the value of $k$ such that the following limit has a finite solution, $L$ such that $L ne 0$:
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos(x)-1)x^k$$
I use the Taylor Series expansions of $e^x$ and $cos(x)$ and simplify the above expression to the following:
$$lim_xrightarrow0 frac-frac14x^6+(frac148-frac112)x^8+(frac1144)x^10x^k$$
Now I have a mental roadblock. For $k<6$ the above limit goes to zero and for $k>6$ this expression should diverge but in my head it goes to zero again.... I am trying to think of a simple example to convince myself but can't. Can someone please help me understand this?
Thanks.
calculus limits taylor-expansion
edited 22 hours ago
Wong Austin
1,097316
asked 22 hours ago
Calum
333
New contributor
Calum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
take for example $k=7$ and divide each term on the numerator by the denominator $x^7$. What do you get?
– Javi
22 hours ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Calculate the value of $k$ such that the following limit has a finite solution, $L$ such that $L ne 0$:
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos(x)-1)x^k$$
I use the Taylor Series expansions of $e^x$ and $cos(x)$ and simplify the above expression to the following:
$$lim_xrightarrow0 frac-frac14x^6+(frac148-frac112)x^8+(frac1144)x^10x^k$$
Now I have a mental roadblock. For $k<6$ the above limit goes to zero and for $k>6$ this expression should diverge but in my head it goes to zero again.... I am trying to think of a simple example to convince myself but can't. Can someone please help me understand this?
Thanks.
calculus limits taylor-expansion
edited 22 hours ago
Wong Austin
1,097316
asked 22 hours ago
Calum
333
New contributor
Calum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Calculate the value of $k$ such that the following limit has a finite solution, $L$ such that $L ne 0$:
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos(x)-1)x^k$$
I use the Taylor Series expansions of $e^x$ and $cos(x)$ and simplify the above expression to the following:
$$lim_xrightarrow0 frac-frac14x^6+(frac148-frac112)x^8+(frac1144)x^10x^k$$
Now I have a mental roadblock. For $k<6$ the above limit goes to zero and for $k>6$ this expression should diverge but in my head it goes to zero again.... I am trying to think of a simple example to convince myself but can't. Can someone please help me understand this?
Thanks.
calculus limits taylor-expansion
calculus limits taylor-expansion
edited 22 hours ago
Wong Austin
1,097316
asked 22 hours ago
Calum
333
New contributor
Calum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 22 hours ago
Wong Austin
1,097316
asked 22 hours ago
Calum
333
New contributor
Calum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 22 hours ago
Wong Austin
1,097316
edited 22 hours ago
Wong Austin
1,097316
edited 22 hours ago
Wong Austin
1,097316
1,097316
asked 22 hours ago
Calum
333
New contributor
Calum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 22 hours ago
Calum
333
asked 22 hours ago
Calum
333
333
New contributor
Calum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Calum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Calum is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
take for example $k=7$ and divide each term on the numerator by the denominator $x^7$. What do you get?
– Javi
22 hours ago
add a comment |Â
take for example $k=7$ and divide each term on the numerator by the denominator $x^7$. What do you get?
– Javi
22 hours ago
take for example $k=7$ and divide each term on the numerator by the denominator $x^7$. What do you get?
– Javi
22 hours ago
take for example $k=7$ and divide each term on the numerator by the denominator $x^7$. What do you get?
– Javi
22 hours ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
4
down vote
accepted
You are almost there
$$lim_xrightarrow0 frac-frac14x^6+(frac148-frac112)x^8+(frac1144)x^10x^k=\
=lim_xrightarrow0 -frac14x^6-k+(frac148-frac112)x^8-k+(frac1144)x^10-k $$
If any of the exponents is negative, then the limit goes to infinity, so you must have $kle 6$.
Also if all exponents are (stricly) positive (this happens iff $k<6$) then the limit goes to zero.
Then you must have $k=6$.
for $k>6$ this expression should diverge but in my head it goes to zero again....
Why? If $k>6$, the first (at least) summand goes to infinity.
The standard recipe for limits with polynomials is: "When $x$ goes to infinity, the highest degree term rules; when $x$ goes to zero, the lowest degree term rules". Here, this suggests to factor the ruling term from the fraction and write it as
$$ x^6-k ( a + b x^2 +cx^4)$$
for some non-zero $a,b,c$. Then, we quickly see that as $xto 0$ the factor in parentheses tends to the constant $a$, and we have the three cases: if $k<6$ the left factor tends to zero, if $k>6$ it tends to infinity, if $k=6$ it tends to one (and hence the limit to $a$).
answered 22 hours ago
leonbloy
38.3k644104
add a comment |Â
up vote
3
down vote
As an alternative derivation, we have that
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos x-1)x^k
=lim_xrightarrow0 frace^x^2-x^2-1x^k-2lim_xrightarrow0 fraccos x-1x^2=-frac12lim_xrightarrow0 frace^x^2-x^2-1x^k-2$$
then recall that
$$e^x^2=1+x^2+frac12x^4+o(x^4)$$
and then for $k=6$
$$lim_xrightarrow0 frace^x^2-x^2-1x^4=lim_xrightarrow0 fracfrac12x^4+o(x^4)x^4=lim_xrightarrow0 fracfrac12+o(1)1=frac12$$
and therefore
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos x-1)x^6=-frac14$$
answered 21 hours ago
gimusi
71.6k73787
You should get rid of those limit signs at the beginning. They clutter up the page, but more importantly, you don't know the limits exist until you prove they do.
– zhw.
15 hours ago
@zhw I agree totally with you, indeed 99.99% of my solutions for limit are without limit signs!
– gimusi
15 hours ago
@zhw. I also don't use limit separation but in latest days I've seen many users solving in that way and so I've decided to change the method a little bit. Moreover here the separation is useful because we can apply Taylor's expansion only to a part using for the other one a standard limit.
– gimusi
15 hours ago
@zhw. Yes of course we are not sure "a priori" whether the separation works or not but "a poteriori" we know that it works.
– gimusi
15 hours ago
But a good proof flows from the unknown to the known. Also, just because some users exhibit bad habits doesn't mean you should. I posted an answer (which I will delete in due time) that is the same as yours but without the extra unneeded stuff.
– zhw.
14 hours ago
 |Â
show 2 more comments
up vote
1
down vote
As an alternative derivation, note
$$frac(e^x^2-x^2-1)(cos x-1)x^k
=frace^x^2-x^2-1x^k-2 fraccos x-1x^2.$$
As is well known, the second fraction on the right $to -1/2.$ As for the first fraction, recall
$$e^x^2=1+x^2+frac12x^4+o(x^4).$$
It follows that
$$frace^x^2-x^2-1x^4tofrac12.$$
Thus taking $k=6,$ we see
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos x-1)x^6=-frac14.$$
answered 14 hours ago
zhw.
67.2k42872
add a comment |Â
up vote
0
down vote
$cos x-1sim -frac12x^2$ and $e^x^2-x^2-1simfrac12x^4$,
So $k=6$ and $L=-frac14$.
answered 21 hours ago
Riemann
3,0481321
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
You are almost there
$$lim_xrightarrow0 frac-frac14x^6+(frac148-frac112)x^8+(frac1144)x^10x^k=\
=lim_xrightarrow0 -frac14x^6-k+(frac148-frac112)x^8-k+(frac1144)x^10-k $$
If any of the exponents is negative, then the limit goes to infinity, so you must have $kle 6$.
Also if all exponents are (stricly) positive (this happens iff $k<6$) then the limit goes to zero.
Then you must have $k=6$.
for $k>6$ this expression should diverge but in my head it goes to zero again....
Why? If $k>6$, the first (at least) summand goes to infinity.
The standard recipe for limits with polynomials is: "When $x$ goes to infinity, the highest degree term rules; when $x$ goes to zero, the lowest degree term rules". Here, this suggests to factor the ruling term from the fraction and write it as
$$ x^6-k ( a + b x^2 +cx^4)$$
for some non-zero $a,b,c$. Then, we quickly see that as $xto 0$ the factor in parentheses tends to the constant $a$, and we have the three cases: if $k<6$ the left factor tends to zero, if $k>6$ it tends to infinity, if $k=6$ it tends to one (and hence the limit to $a$).
answered 22 hours ago
leonbloy
38.3k644104
add a comment |Â
up vote
4
down vote
accepted
You are almost there
$$lim_xrightarrow0 frac-frac14x^6+(frac148-frac112)x^8+(frac1144)x^10x^k=\
=lim_xrightarrow0 -frac14x^6-k+(frac148-frac112)x^8-k+(frac1144)x^10-k $$
If any of the exponents is negative, then the limit goes to infinity, so you must have $kle 6$.
Also if all exponents are (stricly) positive (this happens iff $k<6$) then the limit goes to zero.
Then you must have $k=6$.
for $k>6$ this expression should diverge but in my head it goes to zero again....
Why? If $k>6$, the first (at least) summand goes to infinity.
The standard recipe for limits with polynomials is: "When $x$ goes to infinity, the highest degree term rules; when $x$ goes to zero, the lowest degree term rules". Here, this suggests to factor the ruling term from the fraction and write it as
$$ x^6-k ( a + b x^2 +cx^4)$$
for some non-zero $a,b,c$. Then, we quickly see that as $xto 0$ the factor in parentheses tends to the constant $a$, and we have the three cases: if $k<6$ the left factor tends to zero, if $k>6$ it tends to infinity, if $k=6$ it tends to one (and hence the limit to $a$).
answered 22 hours ago
leonbloy
38.3k644104
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
You are almost there
$$lim_xrightarrow0 frac-frac14x^6+(frac148-frac112)x^8+(frac1144)x^10x^k=\
=lim_xrightarrow0 -frac14x^6-k+(frac148-frac112)x^8-k+(frac1144)x^10-k $$
If any of the exponents is negative, then the limit goes to infinity, so you must have $kle 6$.
Also if all exponents are (stricly) positive (this happens iff $k<6$) then the limit goes to zero.
Then you must have $k=6$.
for $k>6$ this expression should diverge but in my head it goes to zero again....
Why? If $k>6$, the first (at least) summand goes to infinity.
The standard recipe for limits with polynomials is: "When $x$ goes to infinity, the highest degree term rules; when $x$ goes to zero, the lowest degree term rules". Here, this suggests to factor the ruling term from the fraction and write it as
$$ x^6-k ( a + b x^2 +cx^4)$$
for some non-zero $a,b,c$. Then, we quickly see that as $xto 0$ the factor in parentheses tends to the constant $a$, and we have the three cases: if $k<6$ the left factor tends to zero, if $k>6$ it tends to infinity, if $k=6$ it tends to one (and hence the limit to $a$).
answered 22 hours ago
leonbloy
38.3k644104
You are almost there
$$lim_xrightarrow0 frac-frac14x^6+(frac148-frac112)x^8+(frac1144)x^10x^k=\
=lim_xrightarrow0 -frac14x^6-k+(frac148-frac112)x^8-k+(frac1144)x^10-k $$
If any of the exponents is negative, then the limit goes to infinity, so you must have $kle 6$.
Also if all exponents are (stricly) positive (this happens iff $k<6$) then the limit goes to zero.
Then you must have $k=6$.
for $k>6$ this expression should diverge but in my head it goes to zero again....
Why? If $k>6$, the first (at least) summand goes to infinity.
The standard recipe for limits with polynomials is: "When $x$ goes to infinity, the highest degree term rules; when $x$ goes to zero, the lowest degree term rules". Here, this suggests to factor the ruling term from the fraction and write it as
$$ x^6-k ( a + b x^2 +cx^4)$$
for some non-zero $a,b,c$. Then, we quickly see that as $xto 0$ the factor in parentheses tends to the constant $a$, and we have the three cases: if $k<6$ the left factor tends to zero, if $k>6$ it tends to infinity, if $k=6$ it tends to one (and hence the limit to $a$).
answered 22 hours ago
leonbloy
38.3k644104
edited 21 hours ago
answered 22 hours ago
leonbloy
38.3k644104
answered 22 hours ago
leonbloy
38.3k644104
answered 22 hours ago
leonbloy
38.3k644104
38.3k644104
add a comment |Â
add a comment |Â
up vote
3
down vote
As an alternative derivation, we have that
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos x-1)x^k
=lim_xrightarrow0 frace^x^2-x^2-1x^k-2lim_xrightarrow0 fraccos x-1x^2=-frac12lim_xrightarrow0 frace^x^2-x^2-1x^k-2$$
then recall that
$$e^x^2=1+x^2+frac12x^4+o(x^4)$$
and then for $k=6$
$$lim_xrightarrow0 frace^x^2-x^2-1x^4=lim_xrightarrow0 fracfrac12x^4+o(x^4)x^4=lim_xrightarrow0 fracfrac12+o(1)1=frac12$$
and therefore
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos x-1)x^6=-frac14$$
answered 21 hours ago
gimusi
71.6k73787
You should get rid of those limit signs at the beginning. They clutter up the page, but more importantly, you don't know the limits exist until you prove they do.
– zhw.
15 hours ago
@zhw I agree totally with you, indeed 99.99% of my solutions for limit are without limit signs!
– gimusi
15 hours ago
@zhw. I also don't use limit separation but in latest days I've seen many users solving in that way and so I've decided to change the method a little bit. Moreover here the separation is useful because we can apply Taylor's expansion only to a part using for the other one a standard limit.
– gimusi
15 hours ago
@zhw. Yes of course we are not sure "a priori" whether the separation works or not but "a poteriori" we know that it works.
– gimusi
15 hours ago
But a good proof flows from the unknown to the known. Also, just because some users exhibit bad habits doesn't mean you should. I posted an answer (which I will delete in due time) that is the same as yours but without the extra unneeded stuff.
– zhw.
14 hours ago
 |Â
show 2 more comments
up vote
3
down vote
As an alternative derivation, we have that
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos x-1)x^k
=lim_xrightarrow0 frace^x^2-x^2-1x^k-2lim_xrightarrow0 fraccos x-1x^2=-frac12lim_xrightarrow0 frace^x^2-x^2-1x^k-2$$
then recall that
$$e^x^2=1+x^2+frac12x^4+o(x^4)$$
and then for $k=6$
$$lim_xrightarrow0 frace^x^2-x^2-1x^4=lim_xrightarrow0 fracfrac12x^4+o(x^4)x^4=lim_xrightarrow0 fracfrac12+o(1)1=frac12$$
and therefore
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos x-1)x^6=-frac14$$
answered 21 hours ago
gimusi
71.6k73787
You should get rid of those limit signs at the beginning. They clutter up the page, but more importantly, you don't know the limits exist until you prove they do.
– zhw.
15 hours ago
@zhw I agree totally with you, indeed 99.99% of my solutions for limit are without limit signs!
– gimusi
15 hours ago
@zhw. I also don't use limit separation but in latest days I've seen many users solving in that way and so I've decided to change the method a little bit. Moreover here the separation is useful because we can apply Taylor's expansion only to a part using for the other one a standard limit.
– gimusi
15 hours ago
@zhw. Yes of course we are not sure "a priori" whether the separation works or not but "a poteriori" we know that it works.
– gimusi
15 hours ago
But a good proof flows from the unknown to the known. Also, just because some users exhibit bad habits doesn't mean you should. I posted an answer (which I will delete in due time) that is the same as yours but without the extra unneeded stuff.
– zhw.
14 hours ago
 |Â
show 2 more comments
up vote
3
down vote
up vote
3
down vote
As an alternative derivation, we have that
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos x-1)x^k
=lim_xrightarrow0 frace^x^2-x^2-1x^k-2lim_xrightarrow0 fraccos x-1x^2=-frac12lim_xrightarrow0 frace^x^2-x^2-1x^k-2$$
then recall that
$$e^x^2=1+x^2+frac12x^4+o(x^4)$$
and then for $k=6$
$$lim_xrightarrow0 frace^x^2-x^2-1x^4=lim_xrightarrow0 fracfrac12x^4+o(x^4)x^4=lim_xrightarrow0 fracfrac12+o(1)1=frac12$$
and therefore
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos x-1)x^6=-frac14$$
answered 21 hours ago
gimusi
71.6k73787
As an alternative derivation, we have that
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos x-1)x^k
=lim_xrightarrow0 frace^x^2-x^2-1x^k-2lim_xrightarrow0 fraccos x-1x^2=-frac12lim_xrightarrow0 frace^x^2-x^2-1x^k-2$$
then recall that
$$e^x^2=1+x^2+frac12x^4+o(x^4)$$
and then for $k=6$
$$lim_xrightarrow0 frace^x^2-x^2-1x^4=lim_xrightarrow0 fracfrac12x^4+o(x^4)x^4=lim_xrightarrow0 fracfrac12+o(1)1=frac12$$
and therefore
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos x-1)x^6=-frac14$$
answered 21 hours ago
gimusi
71.6k73787
edited 15 hours ago
answered 21 hours ago
gimusi
71.6k73787
answered 21 hours ago
gimusi
71.6k73787
answered 21 hours ago
gimusi
71.6k73787
71.6k73787
You should get rid of those limit signs at the beginning. They clutter up the page, but more importantly, you don't know the limits exist until you prove they do.
– zhw.
15 hours ago
@zhw I agree totally with you, indeed 99.99% of my solutions for limit are without limit signs!
– gimusi
15 hours ago
@zhw. I also don't use limit separation but in latest days I've seen many users solving in that way and so I've decided to change the method a little bit. Moreover here the separation is useful because we can apply Taylor's expansion only to a part using for the other one a standard limit.
– gimusi
15 hours ago
@zhw. Yes of course we are not sure "a priori" whether the separation works or not but "a poteriori" we know that it works.
– gimusi
15 hours ago
But a good proof flows from the unknown to the known. Also, just because some users exhibit bad habits doesn't mean you should. I posted an answer (which I will delete in due time) that is the same as yours but without the extra unneeded stuff.
– zhw.
14 hours ago
 |Â
show 2 more comments
You should get rid of those limit signs at the beginning. They clutter up the page, but more importantly, you don't know the limits exist until you prove they do.
– zhw.
15 hours ago
@zhw I agree totally with you, indeed 99.99% of my solutions for limit are without limit signs!
– gimusi
15 hours ago
@zhw. I also don't use limit separation but in latest days I've seen many users solving in that way and so I've decided to change the method a little bit. Moreover here the separation is useful because we can apply Taylor's expansion only to a part using for the other one a standard limit.
– gimusi
15 hours ago
@zhw. Yes of course we are not sure "a priori" whether the separation works or not but "a poteriori" we know that it works.
– gimusi
15 hours ago
But a good proof flows from the unknown to the known. Also, just because some users exhibit bad habits doesn't mean you should. I posted an answer (which I will delete in due time) that is the same as yours but without the extra unneeded stuff.
– zhw.
14 hours ago
You should get rid of those limit signs at the beginning. They clutter up the page, but more importantly, you don't know the limits exist until you prove they do.
– zhw.
15 hours ago
You should get rid of those limit signs at the beginning. They clutter up the page, but more importantly, you don't know the limits exist until you prove they do.
– zhw.
15 hours ago
@zhw I agree totally with you, indeed 99.99% of my solutions for limit are without limit signs!
– gimusi
15 hours ago
@zhw I agree totally with you, indeed 99.99% of my solutions for limit are without limit signs!
– gimusi
15 hours ago
@zhw. I also don't use limit separation but in latest days I've seen many users solving in that way and so I've decided to change the method a little bit. Moreover here the separation is useful because we can apply Taylor's expansion only to a part using for the other one a standard limit.
– gimusi
15 hours ago
@zhw. I also don't use limit separation but in latest days I've seen many users solving in that way and so I've decided to change the method a little bit. Moreover here the separation is useful because we can apply Taylor's expansion only to a part using for the other one a standard limit.
– gimusi
15 hours ago
@zhw. Yes of course we are not sure "a priori" whether the separation works or not but "a poteriori" we know that it works.
– gimusi
15 hours ago
@zhw. Yes of course we are not sure "a priori" whether the separation works or not but "a poteriori" we know that it works.
– gimusi
15 hours ago
But a good proof flows from the unknown to the known. Also, just because some users exhibit bad habits doesn't mean you should. I posted an answer (which I will delete in due time) that is the same as yours but without the extra unneeded stuff.
– zhw.
14 hours ago
But a good proof flows from the unknown to the known. Also, just because some users exhibit bad habits doesn't mean you should. I posted an answer (which I will delete in due time) that is the same as yours but without the extra unneeded stuff.
– zhw.
14 hours ago
 |Â
show 2 more comments
up vote
1
down vote
As an alternative derivation, note
$$frac(e^x^2-x^2-1)(cos x-1)x^k
=frace^x^2-x^2-1x^k-2 fraccos x-1x^2.$$
As is well known, the second fraction on the right $to -1/2.$ As for the first fraction, recall
$$e^x^2=1+x^2+frac12x^4+o(x^4).$$
It follows that
$$frace^x^2-x^2-1x^4tofrac12.$$
Thus taking $k=6,$ we see
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos x-1)x^6=-frac14.$$
answered 14 hours ago
zhw.
67.2k42872
add a comment |Â
up vote
1
down vote
As an alternative derivation, note
$$frac(e^x^2-x^2-1)(cos x-1)x^k
=frace^x^2-x^2-1x^k-2 fraccos x-1x^2.$$
As is well known, the second fraction on the right $to -1/2.$ As for the first fraction, recall
$$e^x^2=1+x^2+frac12x^4+o(x^4).$$
It follows that
$$frace^x^2-x^2-1x^4tofrac12.$$
Thus taking $k=6,$ we see
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos x-1)x^6=-frac14.$$
answered 14 hours ago
zhw.
67.2k42872
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As an alternative derivation, note
$$frac(e^x^2-x^2-1)(cos x-1)x^k
=frace^x^2-x^2-1x^k-2 fraccos x-1x^2.$$
As is well known, the second fraction on the right $to -1/2.$ As for the first fraction, recall
$$e^x^2=1+x^2+frac12x^4+o(x^4).$$
It follows that
$$frace^x^2-x^2-1x^4tofrac12.$$
Thus taking $k=6,$ we see
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos x-1)x^6=-frac14.$$
answered 14 hours ago
zhw.
67.2k42872
As an alternative derivation, note
$$frac(e^x^2-x^2-1)(cos x-1)x^k
=frace^x^2-x^2-1x^k-2 fraccos x-1x^2.$$
As is well known, the second fraction on the right $to -1/2.$ As for the first fraction, recall
$$e^x^2=1+x^2+frac12x^4+o(x^4).$$
It follows that
$$frace^x^2-x^2-1x^4tofrac12.$$
Thus taking $k=6,$ we see
$$lim_xrightarrow0 frac(e^x^2-x^2-1)(cos x-1)x^6=-frac14.$$
answered 14 hours ago
zhw.
67.2k42872
answered 14 hours ago
zhw.
67.2k42872
answered 14 hours ago
zhw.
67.2k42872
answered 14 hours ago
zhw.
67.2k42872
67.2k42872
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up vote
0
down vote
$cos x-1sim -frac12x^2$ and $e^x^2-x^2-1simfrac12x^4$,
So $k=6$ and $L=-frac14$.
answered 21 hours ago
Riemann
3,0481321
add a comment |Â
up vote
0
down vote
$cos x-1sim -frac12x^2$ and $e^x^2-x^2-1simfrac12x^4$,
So $k=6$ and $L=-frac14$.
answered 21 hours ago
Riemann
3,0481321
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$cos x-1sim -frac12x^2$ and $e^x^2-x^2-1simfrac12x^4$,
So $k=6$ and $L=-frac14$.
answered 21 hours ago
Riemann
3,0481321
$cos x-1sim -frac12x^2$ and $e^x^2-x^2-1simfrac12x^4$,
So $k=6$ and $L=-frac14$.
answered 21 hours ago
Riemann
3,0481321
answered 21 hours ago
Riemann
3,0481321
answered 21 hours ago
Riemann
3,0481321
answered 21 hours ago
Riemann
3,0481321
3,0481321
add a comment |Â
add a comment |Â
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take for example $k=7$ and divide each term on the numerator by the denominator $x^7$. What do you get?
– Javi
22 hours ago