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Limit related problems in differentiation
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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 1 hour ago
GoodDeeds
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asked 1 hour ago
Ramisa Samira
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up vote
1
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 1 hour ago
GoodDeeds
10.2k21335
asked 1 hour ago
Ramisa Samira
91
New contributor
Ramisa Samira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Welcome to Math SE! Your image is barely readable. Since the question is so short, how about type it in MathJax?
– Toby Mak
1 hour ago
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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.
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1 hour ago
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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
1 hour ago
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up vote
1
down vote
favorite
up vote
1
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 1 hour ago
GoodDeeds
10.2k21335
asked 1 hour ago
Ramisa Samira
91
New contributor
Ramisa Samira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 1 hour ago
GoodDeeds
10.2k21335
asked 1 hour ago
Ramisa Samira
91
New contributor
Ramisa Samira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
GoodDeeds
10.2k21335
asked 1 hour ago
Ramisa Samira
91
New contributor
Ramisa Samira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
GoodDeeds
10.2k21335
edited 1 hour ago
GoodDeeds
10.2k21335
edited 1 hour ago
GoodDeeds
10.2k21335
10.2k21335
asked 1 hour ago
Ramisa Samira
91
New contributor
Ramisa Samira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Ramisa Samira
91
asked 1 hour ago
Ramisa Samira
91
91
New contributor
Ramisa Samira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ramisa Samira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ramisa Samira is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Welcome to Math SE! Your image is barely readable. Since the question is so short, how about type it in MathJax?
– Toby Mak
1 hour ago
1
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
1 hour ago
What kind of rules can you use?
– Dr. Sonnhard Graubner
1 hour ago
1
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
1 hour ago
add a comment |Â
1
Welcome to Math SE! Your image is barely readable. Since the question is so short, how about type it in MathJax?
– Toby Mak
1 hour ago
1
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
1 hour ago
What kind of rules can you use?
– Dr. Sonnhard Graubner
1 hour ago
1
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
1 hour ago
1
1
Welcome to Math SE! Your image is barely readable. Since the question is so short, how about type it in MathJax?
– Toby Mak
1 hour ago
Welcome to Math SE! Your image is barely readable. Since the question is so short, how about type it in MathJax?
– Toby Mak
1 hour ago
1
1
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
1 hour ago
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
1 hour ago
What kind of rules can you use?
– Dr. Sonnhard Graubner
1 hour ago
What kind of rules can you use?
– Dr. Sonnhard Graubner
1 hour ago
1
1
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
1 hour ago
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
1 hour ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
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 1 hour ago
gimusi
72.3k73888
add a comment |Â
up vote
2
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 1 hour ago
mechanodroid
24.2k52245
1
@MathOverview Yes. Check that $$ax^2+bx+ c = aleft(x - frac-b+sqrtb^2-4ac2aright)left(x - frac-b-sqrtb^2-4ac2aright)$$
– mechanodroid
1 hour ago
add a comment |Â
up vote
1
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 1 hour ago
MathOverview
8,34442962
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
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 1 hour ago
gimusi
72.3k73888
add a comment |Â
up vote
2
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 1 hour ago
gimusi
72.3k73888
add a comment |Â
up vote
2
down vote
up vote
2
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 1 hour ago
gimusi
72.3k73888
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 1 hour ago
gimusi
72.3k73888
answered 1 hour ago
gimusi
72.3k73888
answered 1 hour ago
gimusi
72.3k73888
answered 1 hour ago
gimusi
72.3k73888
72.3k73888
add a comment |Â
add a comment |Â
up vote
2
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 1 hour ago
mechanodroid
24.2k52245
1
@MathOverview Yes. Check that $$ax^2+bx+ c = aleft(x - frac-b+sqrtb^2-4ac2aright)left(x - frac-b-sqrtb^2-4ac2aright)$$
– mechanodroid
1 hour ago
add a comment |Â
up vote
2
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 1 hour ago
mechanodroid
24.2k52245
1
@MathOverview Yes. Check that $$ax^2+bx+ c = aleft(x - frac-b+sqrtb^2-4ac2aright)left(x - frac-b-sqrtb^2-4ac2aright)$$
– mechanodroid
1 hour ago
add a comment |Â
up vote
2
down vote
up vote
2
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 1 hour ago
mechanodroid
24.2k52245
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 1 hour ago
mechanodroid
24.2k52245
answered 1 hour ago
mechanodroid
24.2k52245
answered 1 hour ago
mechanodroid
24.2k52245
answered 1 hour ago
mechanodroid
24.2k52245
24.2k52245
1
@MathOverview Yes. Check that $$ax^2+bx+ c = aleft(x - frac-b+sqrtb^2-4ac2aright)left(x - frac-b-sqrtb^2-4ac2aright)$$
– mechanodroid
1 hour ago
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
1 hour ago
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
1 hour ago
@MathOverview Yes. Check that $$ax^2+bx+ c = aleft(x - frac-b+sqrtb^2-4ac2aright)left(x - frac-b-sqrtb^2-4ac2aright)$$
– mechanodroid
1 hour ago
add a comment |Â
up vote
1
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 1 hour ago
MathOverview
8,34442962
add a comment |Â
up vote
1
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 1 hour ago
MathOverview
8,34442962
add a comment |Â
up vote
1
down vote
up vote
1
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 1 hour ago
MathOverview
8,34442962
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 1 hour ago
MathOverview
8,34442962
edited 1 hour ago
answered 1 hour ago
MathOverview
8,34442962
answered 1 hour ago
MathOverview
8,34442962
answered 1 hour ago
MathOverview
8,34442962
8,34442962
add a comment |Â
add a comment |Â
Ramisa Samira is a new contributor. Be nice, and check out our Code of Conduct.
Ramisa Samira is a new contributor. Be nice, and check out our Code of Conduct.
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1
Welcome to Math SE! Your image is barely readable. Since the question is so short, how about type it in MathJax?
– Toby Mak
1 hour ago
1
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
1 hour ago
What kind of rules can you use?
– Dr. Sonnhard Graubner
1 hour ago
1
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
1 hour ago