Mixing
Limit based question (wanna know how to recognize such questions in first glance)
$begingroup$
$lim_{xto 0}frac{x-sinx}{(tanx)^3}$
I have its solution,
In the solution, $sinx$ and $tanx$ both are expanded (i think by mclaurin expansion; not sure)
I tried to solve it by L hospital rule but couldn't solve
Is there anyway to recognize such questions in first glance?
limits
asked Jan 11 at 16:48
AkashAkash
786
$endgroup$
add a comment |
$begingroup$
$lim_{xto 0}frac{x-sinx}{(tanx)^3}$
I have its solution,
In the solution, $sinx$ and $tanx$ both are expanded (i think by mclaurin expansion; not sure)
I tried to solve it by L hospital rule but couldn't solve
Is there anyway to recognize such questions in first glance?
limits
asked Jan 11 at 16:48
AkashAkash
786
$endgroup$
$begingroup$
$x-sin(x) sim x^3/6$, $tan(x)^3 sim x^3$...
$endgroup$
– Mindlack
Jan 11 at 16:50
1
$begingroup$
Can you use the rules of L'Hospital?
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 16:55
$begingroup$
Yes i can use L'Hospital
$endgroup$
– Akash
Jan 11 at 17:10
2
$begingroup$
Your parenthetical remark is (I think) the most reliable general method. Sometimes you can recognize a friend like $(sin x)/x$. L'Hopital should be a last resort.
$endgroup$
– Ethan Bolker
Jan 11 at 17:38
add a comment |
$begingroup$
$lim_{xto 0}frac{x-sinx}{(tanx)^3}$
I have its solution,
In the solution, $sinx$ and $tanx$ both are expanded (i think by mclaurin expansion; not sure)
I tried to solve it by L hospital rule but couldn't solve
Is there anyway to recognize such questions in first glance?
limits
asked Jan 11 at 16:48
AkashAkash
786
$endgroup$
$lim_{xto 0}frac{x-sinx}{(tanx)^3}$
I have its solution,
In the solution, $sinx$ and $tanx$ both are expanded (i think by mclaurin expansion; not sure)
I tried to solve it by L hospital rule but couldn't solve
Is there anyway to recognize such questions in first glance?
limits
limits
asked Jan 11 at 16:48
AkashAkash
786
asked Jan 11 at 16:48
AkashAkash
786
asked Jan 11 at 16:48
AkashAkash
786
asked Jan 11 at 16:48
AkashAkash
786
asked Jan 11 at 16:48
AkashAkash
786
786
$begingroup$
$x-sin(x) sim x^3/6$, $tan(x)^3 sim x^3$...
$endgroup$
– Mindlack
Jan 11 at 16:50
1
$begingroup$
Can you use the rules of L'Hospital?
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 16:55
$begingroup$
Yes i can use L'Hospital
$endgroup$
– Akash
Jan 11 at 17:10
2
$begingroup$
Your parenthetical remark is (I think) the most reliable general method. Sometimes you can recognize a friend like $(sin x)/x$. L'Hopital should be a last resort.
$endgroup$
– Ethan Bolker
Jan 11 at 17:38
add a comment |
$begingroup$
$x-sin(x) sim x^3/6$, $tan(x)^3 sim x^3$...
$endgroup$
– Mindlack
Jan 11 at 16:50
1
$begingroup$
Can you use the rules of L'Hospital?
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 16:55
$begingroup$
Yes i can use L'Hospital
$endgroup$
– Akash
Jan 11 at 17:10
2
$begingroup$
Your parenthetical remark is (I think) the most reliable general method. Sometimes you can recognize a friend like $(sin x)/x$. L'Hopital should be a last resort.
$endgroup$
– Ethan Bolker
Jan 11 at 17:38
$begingroup$
$x-sin(x) sim x^3/6$, $tan(x)^3 sim x^3$...
$endgroup$
– Mindlack
Jan 11 at 16:50
$begingroup$
$x-sin(x) sim x^3/6$, $tan(x)^3 sim x^3$...
$endgroup$
– Mindlack
Jan 11 at 16:50
1
1
$begingroup$
Can you use the rules of L'Hospital?
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 16:55
$begingroup$
Can you use the rules of L'Hospital?
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 16:55
$begingroup$
Yes i can use L'Hospital
$endgroup$
– Akash
Jan 11 at 17:10
$begingroup$
Yes i can use L'Hospital
$endgroup$
– Akash
Jan 11 at 17:10
2
2
$begingroup$
Your parenthetical remark is (I think) the most reliable general method. Sometimes you can recognize a friend like $(sin x)/x$. L'Hopital should be a last resort.
$endgroup$
– Ethan Bolker
Jan 11 at 17:38
$begingroup$
Your parenthetical remark is (I think) the most reliable general method. Sometimes you can recognize a friend like $(sin x)/x$. L'Hopital should be a last resort.
$endgroup$
– Ethan Bolker
Jan 11 at 17:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Just for the record, L'Hôpital's Rule also yields a limiting value of $1/6$.
begin{align}
lim_{xto0}frac{x-sin x}{tan^3x}
& = lim_{xto0}frac{1-cos x}{3tan^2xsec^2x} \
& = lim_{xto0}frac{cos^2x-cos^3x}{3tan^2x} \
& = lim_{xto0}frac{(2cos x-3cos^2x)(-sin x)}{6tan xsec^2x} \
& = lim_{xto0}frac12cos^5x-frac134cos^4x \
& = frac16
end{align}
I'm not sure that there's any infallible detection algorithm for such questions, other than recognizing that all the terms have a fairly well-behaved MacLaurin series. Sometimes it might be disguised in a trigonometric expression (though it isn't here), which can be "unlocked" by simplifying it, a little bit like I did in the L'Hôpital's Rule expansion above.
answered Jan 11 at 16:56
Brian TungBrian Tung
25.8k32554
$endgroup$
$begingroup$
Thabk you..i got it
$endgroup$
– Akash
Jan 11 at 17:04
add a comment |
$begingroup$
Use the fundamental limit $$lim_{xto 0}frac {tan x}x=1$$ , then you "get rid" of the tangent and then it is easy with L'Hopital.
edited Jan 11 at 17:35
Thomas Shelby
2,8971421
answered Jan 11 at 17:00
amarius8312amarius8312
1435
$endgroup$
1
$begingroup$
Then i have to multiply and divide the tangent by $x^3$?
$endgroup$
– Akash
Jan 11 at 17:04
$begingroup$
Ohh..i solved it this way, thank you.. (I don't know why they have given a long solution in the book when it can be done this way)
$endgroup$
– Akash
Jan 11 at 17:08
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Just for the record, L'Hôpital's Rule also yields a limiting value of $1/6$.
begin{align}
lim_{xto0}frac{x-sin x}{tan^3x}
& = lim_{xto0}frac{1-cos x}{3tan^2xsec^2x} \
& = lim_{xto0}frac{cos^2x-cos^3x}{3tan^2x} \
& = lim_{xto0}frac{(2cos x-3cos^2x)(-sin x)}{6tan xsec^2x} \
& = lim_{xto0}frac12cos^5x-frac134cos^4x \
& = frac16
end{align}
I'm not sure that there's any infallible detection algorithm for such questions, other than recognizing that all the terms have a fairly well-behaved MacLaurin series. Sometimes it might be disguised in a trigonometric expression (though it isn't here), which can be "unlocked" by simplifying it, a little bit like I did in the L'Hôpital's Rule expansion above.
answered Jan 11 at 16:56
Brian TungBrian Tung
25.8k32554
$endgroup$
$begingroup$
Thabk you..i got it
$endgroup$
– Akash
Jan 11 at 17:04
add a comment |
$begingroup$
Just for the record, L'Hôpital's Rule also yields a limiting value of $1/6$.
begin{align}
lim_{xto0}frac{x-sin x}{tan^3x}
& = lim_{xto0}frac{1-cos x}{3tan^2xsec^2x} \
& = lim_{xto0}frac{cos^2x-cos^3x}{3tan^2x} \
& = lim_{xto0}frac{(2cos x-3cos^2x)(-sin x)}{6tan xsec^2x} \
& = lim_{xto0}frac12cos^5x-frac134cos^4x \
& = frac16
end{align}
I'm not sure that there's any infallible detection algorithm for such questions, other than recognizing that all the terms have a fairly well-behaved MacLaurin series. Sometimes it might be disguised in a trigonometric expression (though it isn't here), which can be "unlocked" by simplifying it, a little bit like I did in the L'Hôpital's Rule expansion above.
answered Jan 11 at 16:56
Brian TungBrian Tung
25.8k32554
$endgroup$
$begingroup$
Thabk you..i got it
$endgroup$
– Akash
Jan 11 at 17:04
add a comment |
$begingroup$
Just for the record, L'Hôpital's Rule also yields a limiting value of $1/6$.
begin{align}
lim_{xto0}frac{x-sin x}{tan^3x}
& = lim_{xto0}frac{1-cos x}{3tan^2xsec^2x} \
& = lim_{xto0}frac{cos^2x-cos^3x}{3tan^2x} \
& = lim_{xto0}frac{(2cos x-3cos^2x)(-sin x)}{6tan xsec^2x} \
& = lim_{xto0}frac12cos^5x-frac134cos^4x \
& = frac16
end{align}
I'm not sure that there's any infallible detection algorithm for such questions, other than recognizing that all the terms have a fairly well-behaved MacLaurin series. Sometimes it might be disguised in a trigonometric expression (though it isn't here), which can be "unlocked" by simplifying it, a little bit like I did in the L'Hôpital's Rule expansion above.
answered Jan 11 at 16:56
Brian TungBrian Tung
25.8k32554
$endgroup$
Just for the record, L'Hôpital's Rule also yields a limiting value of $1/6$.
begin{align}
lim_{xto0}frac{x-sin x}{tan^3x}
& = lim_{xto0}frac{1-cos x}{3tan^2xsec^2x} \
& = lim_{xto0}frac{cos^2x-cos^3x}{3tan^2x} \
& = lim_{xto0}frac{(2cos x-3cos^2x)(-sin x)}{6tan xsec^2x} \
& = lim_{xto0}frac12cos^5x-frac134cos^4x \
& = frac16
end{align}
I'm not sure that there's any infallible detection algorithm for such questions, other than recognizing that all the terms have a fairly well-behaved MacLaurin series. Sometimes it might be disguised in a trigonometric expression (though it isn't here), which can be "unlocked" by simplifying it, a little bit like I did in the L'Hôpital's Rule expansion above.
answered Jan 11 at 16:56
Brian TungBrian Tung
25.8k32554
answered Jan 11 at 16:56
Brian TungBrian Tung
25.8k32554
answered Jan 11 at 16:56
Brian TungBrian Tung
25.8k32554
answered Jan 11 at 16:56
Brian TungBrian Tung
25.8k32554
25.8k32554
$begingroup$
Thabk you..i got it
$endgroup$
– Akash
Jan 11 at 17:04
add a comment |
$begingroup$
Thabk you..i got it
$endgroup$
– Akash
Jan 11 at 17:04
$begingroup$
Thabk you..i got it
$endgroup$
– Akash
Jan 11 at 17:04
$begingroup$
Thabk you..i got it
$endgroup$
– Akash
Jan 11 at 17:04
add a comment |
$begingroup$
Use the fundamental limit $$lim_{xto 0}frac {tan x}x=1$$ , then you "get rid" of the tangent and then it is easy with L'Hopital.
edited Jan 11 at 17:35
Thomas Shelby
2,8971421
answered Jan 11 at 17:00
amarius8312amarius8312
1435
$endgroup$
1
$begingroup$
Then i have to multiply and divide the tangent by $x^3$?
$endgroup$
– Akash
Jan 11 at 17:04
$begingroup$
Ohh..i solved it this way, thank you.. (I don't know why they have given a long solution in the book when it can be done this way)
$endgroup$
– Akash
Jan 11 at 17:08
add a comment |
$begingroup$
Use the fundamental limit $$lim_{xto 0}frac {tan x}x=1$$ , then you "get rid" of the tangent and then it is easy with L'Hopital.
edited Jan 11 at 17:35
Thomas Shelby
2,8971421
answered Jan 11 at 17:00
amarius8312amarius8312
1435
$endgroup$
1
$begingroup$
Then i have to multiply and divide the tangent by $x^3$?
$endgroup$
– Akash
Jan 11 at 17:04
$begingroup$
Ohh..i solved it this way, thank you.. (I don't know why they have given a long solution in the book when it can be done this way)
$endgroup$
– Akash
Jan 11 at 17:08
add a comment |
$begingroup$
Use the fundamental limit $$lim_{xto 0}frac {tan x}x=1$$ , then you "get rid" of the tangent and then it is easy with L'Hopital.
edited Jan 11 at 17:35
Thomas Shelby
2,8971421
answered Jan 11 at 17:00
amarius8312amarius8312
1435
$endgroup$
Use the fundamental limit $$lim_{xto 0}frac {tan x}x=1$$ , then you "get rid" of the tangent and then it is easy with L'Hopital.
edited Jan 11 at 17:35
Thomas Shelby
2,8971421
answered Jan 11 at 17:00
amarius8312amarius8312
1435
edited Jan 11 at 17:35
Thomas Shelby
2,8971421
edited Jan 11 at 17:35
Thomas Shelby
2,8971421
edited Jan 11 at 17:35
Thomas Shelby
2,8971421
2,8971421
answered Jan 11 at 17:00
amarius8312amarius8312
1435
answered Jan 11 at 17:00
amarius8312amarius8312
1435
answered Jan 11 at 17:00
amarius8312amarius8312
1435
1435
1
$begingroup$
Then i have to multiply and divide the tangent by $x^3$?
$endgroup$
– Akash
Jan 11 at 17:04
$begingroup$
Ohh..i solved it this way, thank you.. (I don't know why they have given a long solution in the book when it can be done this way)
$endgroup$
– Akash
Jan 11 at 17:08
add a comment |
1
$begingroup$
Then i have to multiply and divide the tangent by $x^3$?
$endgroup$
– Akash
Jan 11 at 17:04
$begingroup$
Ohh..i solved it this way, thank you.. (I don't know why they have given a long solution in the book when it can be done this way)
$endgroup$
– Akash
Jan 11 at 17:08
1
1
$begingroup$
Then i have to multiply and divide the tangent by $x^3$?
$endgroup$
– Akash
Jan 11 at 17:04
$begingroup$
Then i have to multiply and divide the tangent by $x^3$?
$endgroup$
– Akash
Jan 11 at 17:04
$begingroup$
Ohh..i solved it this way, thank you.. (I don't know why they have given a long solution in the book when it can be done this way)
$endgroup$
– Akash
Jan 11 at 17:08
$begingroup$
Ohh..i solved it this way, thank you.. (I don't know why they have given a long solution in the book when it can be done this way)
$endgroup$
– Akash
Jan 11 at 17:08
add a comment |
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$begingroup$
$x-sin(x) sim x^3/6$, $tan(x)^3 sim x^3$...
$endgroup$
– Mindlack
Jan 11 at 16:50
1
$begingroup$
Can you use the rules of L'Hospital?
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 16:55
$begingroup$
Yes i can use L'Hospital
$endgroup$
– Akash
Jan 11 at 17:10
2
$begingroup$
Your parenthetical remark is (I think) the most reliable general method. Sometimes you can recognize a friend like $(sin x)/x$. L'Hopital should be a last resort.
$endgroup$
– Ethan Bolker
Jan 11 at 17:38