Finding the limit $frac {sin x - arctan x}{x^2 ln x}$ as $x rightarrow 0$ in Petrovic
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2
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Finding the limit:
$$lim_{x to 0}frac {sin x - arctan x}{x^2 ln x}$$
My questions:
1- I think the question should be corrected to as $x rightarrow 0^+$, because of the domain of $ln x$ ...... am I correct?
2- I applied L`hopital 2 times and after the second time it gaves me $0/-infty $ which is $0$ .... am I correct?
calculus limits
edited 2 hours ago
PradyumanDixit
774214
asked 3 hours ago
hopefully
1868
add a comment |
up vote
2
down vote
favorite
Finding the limit:
$$lim_{x to 0}frac {sin x - arctan x}{x^2 ln x}$$
My questions:
1- I think the question should be corrected to as $x rightarrow 0^+$, because of the domain of $ln x$ ...... am I correct?
2- I applied L`hopital 2 times and after the second time it gaves me $0/-infty $ which is $0$ .... am I correct?
calculus limits
edited 2 hours ago
PradyumanDixit
774214
asked 3 hours ago
hopefully
1868
1
Your observations are both correct.
– Kavi Rama Murthy
3 hours ago
1
Yes to both. There are other methods too, such as power series, to show the limit is 0.
– DanielWainfleet
2 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Finding the limit:
$$lim_{x to 0}frac {sin x - arctan x}{x^2 ln x}$$
My questions:
1- I think the question should be corrected to as $x rightarrow 0^+$, because of the domain of $ln x$ ...... am I correct?
2- I applied L`hopital 2 times and after the second time it gaves me $0/-infty $ which is $0$ .... am I correct?
calculus limits
edited 2 hours ago
PradyumanDixit
774214
asked 3 hours ago
hopefully
1868
Finding the limit:
$$lim_{x to 0}frac {sin x - arctan x}{x^2 ln x}$$
My questions:
1- I think the question should be corrected to as $x rightarrow 0^+$, because of the domain of $ln x$ ...... am I correct?
2- I applied L`hopital 2 times and after the second time it gaves me $0/-infty $ which is $0$ .... am I correct?
calculus limits
calculus limits
edited 2 hours ago
PradyumanDixit
774214
asked 3 hours ago
hopefully
1868
edited 2 hours ago
PradyumanDixit
774214
asked 3 hours ago
hopefully
1868
edited 2 hours ago
PradyumanDixit
774214
edited 2 hours ago
PradyumanDixit
774214
edited 2 hours ago
PradyumanDixit
774214
774214
asked 3 hours ago
hopefully
1868
asked 3 hours ago
hopefully
1868
asked 3 hours ago
hopefully
1868
1868
1
Your observations are both correct.
– Kavi Rama Murthy
3 hours ago
1
Yes to both. There are other methods too, such as power series, to show the limit is 0.
– DanielWainfleet
2 hours ago
add a comment |
1
Your observations are both correct.
– Kavi Rama Murthy
3 hours ago
1
Yes to both. There are other methods too, such as power series, to show the limit is 0.
– DanielWainfleet
2 hours ago
1
1
Your observations are both correct.
– Kavi Rama Murthy
3 hours ago
Your observations are both correct.
– Kavi Rama Murthy
3 hours ago
1
1
Yes to both. There are other methods too, such as power series, to show the limit is 0.
– DanielWainfleet
2 hours ago
Yes to both. There are other methods too, such as power series, to show the limit is 0.
– DanielWainfleet
2 hours ago
add a comment |
2 Answers
2
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oldest
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up vote
1
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accepted
Because of the domain of $ln{x}$ the limit should approach as $x rightarrow 0^{+} $
Solving the limit, on seeing the numerator the first thought is expansion but $ln{x}$ does not have an easy expansion, so go for double L'hôspital. This gives
$$ lim_{x rightarrow 0^{+}}frac{-sin{x} + frac{2x}{{1+x}^{2}}}{2ln{x}+3} =0$$
answered 2 hours ago
Lakshya Sinha
364
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Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
1
down vote
Another way:
$$dfrac{sin x-arctan x}{x^2ln x}=left(underbrace{dfrac{sin x-x}{x^3}}-underbrace{dfrac{arctan x-x}{x^3}}right)dfrac x{ln x}$$
Using Are all limits solvable without L'Hôpital Rule or Series Expansion, the terms with underbrace have finite limits.
$$lim_{xto0^+}dfrac x{ln x}=dfrac0{-infty}=0$$
answered 2 hours ago
lab bhattacharjee
219k14153268
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Because of the domain of $ln{x}$ the limit should approach as $x rightarrow 0^{+} $
Solving the limit, on seeing the numerator the first thought is expansion but $ln{x}$ does not have an easy expansion, so go for double L'hôspital. This gives
$$ lim_{x rightarrow 0^{+}}frac{-sin{x} + frac{2x}{{1+x}^{2}}}{2ln{x}+3} =0$$
answered 2 hours ago
Lakshya Sinha
364
New contributor
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
accepted
Because of the domain of $ln{x}$ the limit should approach as $x rightarrow 0^{+} $
Solving the limit, on seeing the numerator the first thought is expansion but $ln{x}$ does not have an easy expansion, so go for double L'hôspital. This gives
$$ lim_{x rightarrow 0^{+}}frac{-sin{x} + frac{2x}{{1+x}^{2}}}{2ln{x}+3} =0$$
answered 2 hours ago
Lakshya Sinha
364
New contributor
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Because of the domain of $ln{x}$ the limit should approach as $x rightarrow 0^{+} $
Solving the limit, on seeing the numerator the first thought is expansion but $ln{x}$ does not have an easy expansion, so go for double L'hôspital. This gives
$$ lim_{x rightarrow 0^{+}}frac{-sin{x} + frac{2x}{{1+x}^{2}}}{2ln{x}+3} =0$$
answered 2 hours ago
Lakshya Sinha
364
New contributor
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Because of the domain of $ln{x}$ the limit should approach as $x rightarrow 0^{+} $
Solving the limit, on seeing the numerator the first thought is expansion but $ln{x}$ does not have an easy expansion, so go for double L'hôspital. This gives
$$ lim_{x rightarrow 0^{+}}frac{-sin{x} + frac{2x}{{1+x}^{2}}}{2ln{x}+3} =0$$
answered 2 hours ago
Lakshya Sinha
364
New contributor
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Lakshya Sinha
364
New contributor
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Lakshya Sinha
364
answered 2 hours ago
Lakshya Sinha
364
364
New contributor
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Lakshya Sinha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
up vote
1
down vote
Another way:
$$dfrac{sin x-arctan x}{x^2ln x}=left(underbrace{dfrac{sin x-x}{x^3}}-underbrace{dfrac{arctan x-x}{x^3}}right)dfrac x{ln x}$$
Using Are all limits solvable without L'Hôpital Rule or Series Expansion, the terms with underbrace have finite limits.
$$lim_{xto0^+}dfrac x{ln x}=dfrac0{-infty}=0$$
answered 2 hours ago
lab bhattacharjee
219k14153268
add a comment |
up vote
1
down vote
Another way:
$$dfrac{sin x-arctan x}{x^2ln x}=left(underbrace{dfrac{sin x-x}{x^3}}-underbrace{dfrac{arctan x-x}{x^3}}right)dfrac x{ln x}$$
Using Are all limits solvable without L'Hôpital Rule or Series Expansion, the terms with underbrace have finite limits.
$$lim_{xto0^+}dfrac x{ln x}=dfrac0{-infty}=0$$
answered 2 hours ago
lab bhattacharjee
219k14153268
add a comment |
up vote
1
down vote
up vote
1
down vote
Another way:
$$dfrac{sin x-arctan x}{x^2ln x}=left(underbrace{dfrac{sin x-x}{x^3}}-underbrace{dfrac{arctan x-x}{x^3}}right)dfrac x{ln x}$$
Using Are all limits solvable without L'Hôpital Rule or Series Expansion, the terms with underbrace have finite limits.
$$lim_{xto0^+}dfrac x{ln x}=dfrac0{-infty}=0$$
answered 2 hours ago
lab bhattacharjee
219k14153268
Another way:
$$dfrac{sin x-arctan x}{x^2ln x}=left(underbrace{dfrac{sin x-x}{x^3}}-underbrace{dfrac{arctan x-x}{x^3}}right)dfrac x{ln x}$$
Using Are all limits solvable without L'Hôpital Rule or Series Expansion, the terms with underbrace have finite limits.
$$lim_{xto0^+}dfrac x{ln x}=dfrac0{-infty}=0$$
answered 2 hours ago
lab bhattacharjee
219k14153268
answered 2 hours ago
lab bhattacharjee
219k14153268
answered 2 hours ago
lab bhattacharjee
219k14153268
answered 2 hours ago
lab bhattacharjee
219k14153268
219k14153268
add a comment |
add a comment |
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1
Your observations are both correct.
– Kavi Rama Murthy
3 hours ago
1
Yes to both. There are other methods too, such as power series, to show the limit is 0.
– DanielWainfleet
2 hours ago