Limit related problems in differentiation [closed]
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Suppose $alpha$ and $beta$ are two roots of the equation $ax^2+bx+c=0$. Find
$$lim_xtoalphafrac1-cos(ax^2+bx+c)(x-alpha)^2$$
Please help me to solve this calculus problem
calculus
edited Sep 16 at 12:00
GoodDeeds
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asked Sep 16 at 11:57
Ramisa Samira
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closed as off-topic by Gibbs, Did, José Carlos Santos, Math_QED, Micah Sep 16 at 20:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РGibbs, Did, Jos̩ Carlos Santos, Math_QED, Micah
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up vote
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down vote
favorite
Suppose $alpha$ and $beta$ are two roots of the equation $ax^2+bx+c=0$. Find
$$lim_xtoalphafrac1-cos(ax^2+bx+c)(x-alpha)^2$$
Please help me to solve this calculus problem
calculus
edited Sep 16 at 12:00
GoodDeeds
10.2k21335
asked Sep 16 at 11:57
Ramisa Samira
201
closed as off-topic by Gibbs, Did, José Carlos Santos, Math_QED, Micah Sep 16 at 20:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РGibbs, Did, Jos̩ Carlos Santos, Math_QED, Micah
2
Welcome to Math SE! Your image is barely readable. Since the question is so short, how about type it in MathJax?
– Toby Mak
Sep 16 at 11:58
3
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Sep 16 at 12:01
What kind of rules can you use?
– Dr. Sonnhard Graubner
Sep 16 at 12:07
2
Instead of suggesting an edit from a different account, please use your original account and click on "edit" to add details to your question. As it stands, it is not possible if the suggested edit is by you or by someone else pretending to be you. On a side note, using your original account, you can add comments to your own posts irrespective of reputation.
– GoodDeeds
Sep 16 at 12:28
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose $alpha$ and $beta$ are two roots of the equation $ax^2+bx+c=0$. Find
$$lim_xtoalphafrac1-cos(ax^2+bx+c)(x-alpha)^2$$
Please help me to solve this calculus problem
calculus
edited Sep 16 at 12:00
GoodDeeds
10.2k21335
asked Sep 16 at 11:57
Ramisa Samira
201
Suppose $alpha$ and $beta$ are two roots of the equation $ax^2+bx+c=0$. Find
$$lim_xtoalphafrac1-cos(ax^2+bx+c)(x-alpha)^2$$
Please help me to solve this calculus problem
calculus
calculus
edited Sep 16 at 12:00
GoodDeeds
10.2k21335
asked Sep 16 at 11:57
Ramisa Samira
201
edited Sep 16 at 12:00
GoodDeeds
10.2k21335
asked Sep 16 at 11:57
Ramisa Samira
201
edited Sep 16 at 12:00
GoodDeeds
10.2k21335
edited Sep 16 at 12:00
GoodDeeds
10.2k21335
edited Sep 16 at 12:00
GoodDeeds
10.2k21335
10.2k21335
asked Sep 16 at 11:57
Ramisa Samira
201
asked Sep 16 at 11:57
Ramisa Samira
201
asked Sep 16 at 11:57
Ramisa Samira
201
201
closed as off-topic by Gibbs, Did, José Carlos Santos, Math_QED, Micah Sep 16 at 20:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РGibbs, Did, Jos̩ Carlos Santos, Math_QED, Micah
closed as off-topic by Gibbs, Did, José Carlos Santos, Math_QED, Micah Sep 16 at 20:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РGibbs, Did, Jos̩ Carlos Santos, Math_QED, Micah
2
Welcome to Math SE! Your image is barely readable. Since the question is so short, how about type it in MathJax?
– Toby Mak
Sep 16 at 11:58
3
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Sep 16 at 12:01
What kind of rules can you use?
– Dr. Sonnhard Graubner
Sep 16 at 12:07
2
Instead of suggesting an edit from a different account, please use your original account and click on "edit" to add details to your question. As it stands, it is not possible if the suggested edit is by you or by someone else pretending to be you. On a side note, using your original account, you can add comments to your own posts irrespective of reputation.
– GoodDeeds
Sep 16 at 12:28
add a comment |Â
2
Welcome to Math SE! Your image is barely readable. Since the question is so short, how about type it in MathJax?
– Toby Mak
Sep 16 at 11:58
3
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Sep 16 at 12:01
What kind of rules can you use?
– Dr. Sonnhard Graubner
Sep 16 at 12:07
2
Instead of suggesting an edit from a different account, please use your original account and click on "edit" to add details to your question. As it stands, it is not possible if the suggested edit is by you or by someone else pretending to be you. On a side note, using your original account, you can add comments to your own posts irrespective of reputation.
– GoodDeeds
Sep 16 at 12:28
2
2
Welcome to Math SE! Your image is barely readable. Since the question is so short, how about type it in MathJax?
– Toby Mak
Sep 16 at 11:58
Welcome to Math SE! Your image is barely readable. Since the question is so short, how about type it in MathJax?
– Toby Mak
Sep 16 at 11:58
3
3
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Sep 16 at 12:01
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Sep 16 at 12:01
What kind of rules can you use?
– Dr. Sonnhard Graubner
Sep 16 at 12:07
What kind of rules can you use?
– Dr. Sonnhard Graubner
Sep 16 at 12:07
2
2
Instead of suggesting an edit from a different account, please use your original account and click on "edit" to add details to your question. As it stands, it is not possible if the suggested edit is by you or by someone else pretending to be you. On a side note, using your original account, you can add comments to your own posts irrespective of reputation.
– GoodDeeds
Sep 16 at 12:28
Instead of suggesting an edit from a different account, please use your original account and click on "edit" to add details to your question. As it stands, it is not possible if the suggested edit is by you or by someone else pretending to be you. On a side note, using your original account, you can add comments to your own posts irrespective of reputation.
– GoodDeeds
Sep 16 at 12:28
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
6
down vote
HINT
We have
$$frac1-cos(ax^2+bx+c)(x-alpha)^2=frac1-cos[a(x-alpha)(x-beta)]a^2(x-alpha)^2(x-beta)^2a^2(x-beta)^2$$
then refer to standard limit as $tto 0$
$$frac1-cos tt^2to frac12$$
answered Sep 16 at 12:18
gimusi
1
add a comment |Â
up vote
3
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Recall that $lim_t to 0 frac1-cos tt^2 = frac12$.
We have
$$frac1-cos(ax^2+bx+c)(x-alpha)^2 = frac1-cos[a(x-alpha)(x-beta)](x-alpha)^2 = frac1-cos[a(x-alpha)(x-beta)][a(x-alpha)(x-beta)]^2cdot a^2(x-beta)^2 xrightarrowxtoalpha frac12a^2(alpha-beta)^2$$
answered Sep 16 at 12:19
mechanodroid
24.8k62245
1
@MathOverview Yes. Check that $$ax^2+bx+ c = aleft(x - frac-b+sqrtb^2-4ac2aright)left(x - frac-b-sqrtb^2-4ac2aright)$$
– mechanodroid
Sep 16 at 12:34
add a comment |Â
up vote
2
down vote
Hint. Use the Taylor series of
$$
cos(a(x-alpha)(x+beta))
$$
around $alpha$. That is
$$
cos(a(x-alpha)(x+beta))= 1-frac12(alpha-beta)^2(x-alpha)^2a^2+o((x-alpha)^3)
$$
answered Sep 16 at 12:25
MathOverview
8,40442962
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
HINT
We have
$$frac1-cos(ax^2+bx+c)(x-alpha)^2=frac1-cos[a(x-alpha)(x-beta)]a^2(x-alpha)^2(x-beta)^2a^2(x-beta)^2$$
then refer to standard limit as $tto 0$
$$frac1-cos tt^2to frac12$$
answered Sep 16 at 12:18
gimusi
1
add a comment |Â
up vote
6
down vote
HINT
We have
$$frac1-cos(ax^2+bx+c)(x-alpha)^2=frac1-cos[a(x-alpha)(x-beta)]a^2(x-alpha)^2(x-beta)^2a^2(x-beta)^2$$
then refer to standard limit as $tto 0$
$$frac1-cos tt^2to frac12$$
answered Sep 16 at 12:18
gimusi
1
add a comment |Â
up vote
6
down vote
up vote
6
down vote
HINT
We have
$$frac1-cos(ax^2+bx+c)(x-alpha)^2=frac1-cos[a(x-alpha)(x-beta)]a^2(x-alpha)^2(x-beta)^2a^2(x-beta)^2$$
then refer to standard limit as $tto 0$
$$frac1-cos tt^2to frac12$$
answered Sep 16 at 12:18
gimusi
1
HINT
We have
$$frac1-cos(ax^2+bx+c)(x-alpha)^2=frac1-cos[a(x-alpha)(x-beta)]a^2(x-alpha)^2(x-beta)^2a^2(x-beta)^2$$
then refer to standard limit as $tto 0$
$$frac1-cos tt^2to frac12$$
answered Sep 16 at 12:18
gimusi
1
answered Sep 16 at 12:18
gimusi
1
answered Sep 16 at 12:18
gimusi
1
answered Sep 16 at 12:18
gimusi
1
1
add a comment |Â
add a comment |Â
up vote
3
down vote
Recall that $lim_t to 0 frac1-cos tt^2 = frac12$.
We have
$$frac1-cos(ax^2+bx+c)(x-alpha)^2 = frac1-cos[a(x-alpha)(x-beta)](x-alpha)^2 = frac1-cos[a(x-alpha)(x-beta)][a(x-alpha)(x-beta)]^2cdot a^2(x-beta)^2 xrightarrowxtoalpha frac12a^2(alpha-beta)^2$$
answered Sep 16 at 12:19
mechanodroid
24.8k62245
1
@MathOverview Yes. Check that $$ax^2+bx+ c = aleft(x - frac-b+sqrtb^2-4ac2aright)left(x - frac-b-sqrtb^2-4ac2aright)$$
– mechanodroid
Sep 16 at 12:34
add a comment |Â
up vote
3
down vote
Recall that $lim_t to 0 frac1-cos tt^2 = frac12$.
We have
$$frac1-cos(ax^2+bx+c)(x-alpha)^2 = frac1-cos[a(x-alpha)(x-beta)](x-alpha)^2 = frac1-cos[a(x-alpha)(x-beta)][a(x-alpha)(x-beta)]^2cdot a^2(x-beta)^2 xrightarrowxtoalpha frac12a^2(alpha-beta)^2$$
answered Sep 16 at 12:19
mechanodroid
24.8k62245
1
@MathOverview Yes. Check that $$ax^2+bx+ c = aleft(x - frac-b+sqrtb^2-4ac2aright)left(x - frac-b-sqrtb^2-4ac2aright)$$
– mechanodroid
Sep 16 at 12:34
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Recall that $lim_t to 0 frac1-cos tt^2 = frac12$.
We have
$$frac1-cos(ax^2+bx+c)(x-alpha)^2 = frac1-cos[a(x-alpha)(x-beta)](x-alpha)^2 = frac1-cos[a(x-alpha)(x-beta)][a(x-alpha)(x-beta)]^2cdot a^2(x-beta)^2 xrightarrowxtoalpha frac12a^2(alpha-beta)^2$$
answered Sep 16 at 12:19
mechanodroid
24.8k62245
Recall that $lim_t to 0 frac1-cos tt^2 = frac12$.
We have
$$frac1-cos(ax^2+bx+c)(x-alpha)^2 = frac1-cos[a(x-alpha)(x-beta)](x-alpha)^2 = frac1-cos[a(x-alpha)(x-beta)][a(x-alpha)(x-beta)]^2cdot a^2(x-beta)^2 xrightarrowxtoalpha frac12a^2(alpha-beta)^2$$
answered Sep 16 at 12:19
mechanodroid
24.8k62245
answered Sep 16 at 12:19
mechanodroid
24.8k62245
answered Sep 16 at 12:19
mechanodroid
24.8k62245
answered Sep 16 at 12:19
mechanodroid
24.8k62245
24.8k62245
1
@MathOverview Yes. Check that $$ax^2+bx+ c = aleft(x - frac-b+sqrtb^2-4ac2aright)left(x - frac-b-sqrtb^2-4ac2aright)$$
– mechanodroid
Sep 16 at 12:34
add a comment |Â
1
@MathOverview Yes. Check that $$ax^2+bx+ c = aleft(x - frac-b+sqrtb^2-4ac2aright)left(x - frac-b-sqrtb^2-4ac2aright)$$
– mechanodroid
Sep 16 at 12:34
1
1
@MathOverview Yes. Check that $$ax^2+bx+ c = aleft(x - frac-b+sqrtb^2-4ac2aright)left(x - frac-b-sqrtb^2-4ac2aright)$$
– mechanodroid
Sep 16 at 12:34
@MathOverview Yes. Check that $$ax^2+bx+ c = aleft(x - frac-b+sqrtb^2-4ac2aright)left(x - frac-b-sqrtb^2-4ac2aright)$$
– mechanodroid
Sep 16 at 12:34
add a comment |Â
up vote
2
down vote
Hint. Use the Taylor series of
$$
cos(a(x-alpha)(x+beta))
$$
around $alpha$. That is
$$
cos(a(x-alpha)(x+beta))= 1-frac12(alpha-beta)^2(x-alpha)^2a^2+o((x-alpha)^3)
$$
answered Sep 16 at 12:25
MathOverview
8,40442962
add a comment |Â
up vote
2
down vote
Hint. Use the Taylor series of
$$
cos(a(x-alpha)(x+beta))
$$
around $alpha$. That is
$$
cos(a(x-alpha)(x+beta))= 1-frac12(alpha-beta)^2(x-alpha)^2a^2+o((x-alpha)^3)
$$
answered Sep 16 at 12:25
MathOverview
8,40442962
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint. Use the Taylor series of
$$
cos(a(x-alpha)(x+beta))
$$
around $alpha$. That is
$$
cos(a(x-alpha)(x+beta))= 1-frac12(alpha-beta)^2(x-alpha)^2a^2+o((x-alpha)^3)
$$
answered Sep 16 at 12:25
MathOverview
8,40442962
Hint. Use the Taylor series of
$$
cos(a(x-alpha)(x+beta))
$$
around $alpha$. That is
$$
cos(a(x-alpha)(x+beta))= 1-frac12(alpha-beta)^2(x-alpha)^2a^2+o((x-alpha)^3)
$$
answered Sep 16 at 12:25
MathOverview
8,40442962
edited Sep 16 at 12:35
answered Sep 16 at 12:25
MathOverview
8,40442962
answered Sep 16 at 12:25
MathOverview
8,40442962
answered Sep 16 at 12:25
MathOverview
8,40442962
8,40442962
add a comment |Â
add a comment |Â
2
Welcome to Math SE! Your image is barely readable. Since the question is so short, how about type it in MathJax?
– Toby Mak
Sep 16 at 11:58
3
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Sep 16 at 12:01
What kind of rules can you use?
– Dr. Sonnhard Graubner
Sep 16 at 12:07
2
Instead of suggesting an edit from a different account, please use your original account and click on "edit" to add details to your question. As it stands, it is not possible if the suggested edit is by you or by someone else pretending to be you. On a side note, using your original account, you can add comments to your own posts irrespective of reputation.
– GoodDeeds
Sep 16 at 12:28