Mixing
How can I claim a one-sided limit doesn't exist?
$begingroup$
I have to find the limit $$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}} $$
Now, I tried using L'Hôpital's rule, but it doesn't lead anywhere.
Manually trying to convert the functions to another form doesn't seem to go anywhere either, so I determined that the limit must be undefined.
However, I cannot prove it. What can I do?
real-analysis limits logarithms proof-explanation limits-without-lhopital
edited Feb 28 at 7:46
Michael Rozenberg
109k1896201
asked Jan 30 at 16:21
sddssdds
296
$endgroup$
add a comment |
$begingroup$
I have to find the limit $$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}} $$
Now, I tried using L'Hôpital's rule, but it doesn't lead anywhere.
Manually trying to convert the functions to another form doesn't seem to go anywhere either, so I determined that the limit must be undefined.
However, I cannot prove it. What can I do?
real-analysis limits logarithms proof-explanation limits-without-lhopital
edited Feb 28 at 7:46
Michael Rozenberg
109k1896201
asked Jan 30 at 16:21
sddssdds
296
$endgroup$
2
$begingroup$
Using Taylor, the numerator is $2xcdot x+o(x^3)$ and the limit is $0$.
$endgroup$
– Yves Daoust
Jan 30 at 16:37
$begingroup$
We haven't been taught about Taylor yet, I should add this to my question. However, thanks for the answer, i'll research Taylor on my own.
$endgroup$
– sdds
Jan 30 at 16:40
$begingroup$
To answer the question in the title, see for example here
$endgroup$
– BlueRaja - Danny Pflughoeft
Jan 30 at 21:37
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Clement C.
Feb 3 at 23:26
add a comment |
$begingroup$
I have to find the limit $$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}} $$
Now, I tried using L'Hôpital's rule, but it doesn't lead anywhere.
Manually trying to convert the functions to another form doesn't seem to go anywhere either, so I determined that the limit must be undefined.
However, I cannot prove it. What can I do?
real-analysis limits logarithms proof-explanation limits-without-lhopital
edited Feb 28 at 7:46
Michael Rozenberg
109k1896201
asked Jan 30 at 16:21
sddssdds
296
$endgroup$
I have to find the limit $$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}} $$
Now, I tried using L'Hôpital's rule, but it doesn't lead anywhere.
Manually trying to convert the functions to another form doesn't seem to go anywhere either, so I determined that the limit must be undefined.
However, I cannot prove it. What can I do?
real-analysis limits logarithms proof-explanation limits-without-lhopital
real-analysis limits logarithms proof-explanation limits-without-lhopital
edited Feb 28 at 7:46
Michael Rozenberg
109k1896201
asked Jan 30 at 16:21
sddssdds
296
edited Feb 28 at 7:46
Michael Rozenberg
109k1896201
asked Jan 30 at 16:21
sddssdds
296
edited Feb 28 at 7:46
Michael Rozenberg
109k1896201
edited Feb 28 at 7:46
Michael Rozenberg
109k1896201
edited Feb 28 at 7:46
Michael Rozenberg
109k1896201
109k1896201
asked Jan 30 at 16:21
sddssdds
296
asked Jan 30 at 16:21
sddssdds
296
asked Jan 30 at 16:21
sddssdds
296
296
2
$begingroup$
Using Taylor, the numerator is $2xcdot x+o(x^3)$ and the limit is $0$.
$endgroup$
– Yves Daoust
Jan 30 at 16:37
$begingroup$
We haven't been taught about Taylor yet, I should add this to my question. However, thanks for the answer, i'll research Taylor on my own.
$endgroup$
– sdds
Jan 30 at 16:40
$begingroup$
To answer the question in the title, see for example here
$endgroup$
– BlueRaja - Danny Pflughoeft
Jan 30 at 21:37
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Clement C.
Feb 3 at 23:26
add a comment |
2
$begingroup$
Using Taylor, the numerator is $2xcdot x+o(x^3)$ and the limit is $0$.
$endgroup$
– Yves Daoust
Jan 30 at 16:37
$begingroup$
We haven't been taught about Taylor yet, I should add this to my question. However, thanks for the answer, i'll research Taylor on my own.
$endgroup$
– sdds
Jan 30 at 16:40
$begingroup$
To answer the question in the title, see for example here
$endgroup$
– BlueRaja - Danny Pflughoeft
Jan 30 at 21:37
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Clement C.
Feb 3 at 23:26
2
2
$begingroup$
Using Taylor, the numerator is $2xcdot x+o(x^3)$ and the limit is $0$.
$endgroup$
– Yves Daoust
Jan 30 at 16:37
$begingroup$
Using Taylor, the numerator is $2xcdot x+o(x^3)$ and the limit is $0$.
$endgroup$
– Yves Daoust
Jan 30 at 16:37
$begingroup$
We haven't been taught about Taylor yet, I should add this to my question. However, thanks for the answer, i'll research Taylor on my own.
$endgroup$
– sdds
Jan 30 at 16:40
$begingroup$
We haven't been taught about Taylor yet, I should add this to my question. However, thanks for the answer, i'll research Taylor on my own.
$endgroup$
– sdds
Jan 30 at 16:40
$begingroup$
To answer the question in the title, see for example here
$endgroup$
– BlueRaja - Danny Pflughoeft
Jan 30 at 21:37
$begingroup$
To answer the question in the title, see for example here
$endgroup$
– BlueRaja - Danny Pflughoeft
Jan 30 at 21:37
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Clement C.
Feb 3 at 23:26
$begingroup$
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
$endgroup$
– Clement C.
Feb 3 at 23:26
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Hint:
Use the facts that (1) $lim_{xto 0} frac{ln(1+2x)}{2x} = 1$, (2) $lim_{xto 0} frac{sin x}{x} = 1$, and (3) $lim_{xto 0^+} frac{x^2}{sqrt{x^3}} = 0$ to show the limit is $2cdot 1cdot 0 = 0$.
To do so, rewrite, for $x>0$,
$$
frac{ln(1+2x)sin x}{sqrt{x^3}}=2cdot frac{ln(1+2x)}{2x}cdotfrac{sin x}{x}cdotfrac{x^2}{sqrt{x^3}}
$$
answered Jan 30 at 16:24
Clement C.Clement C.
51k34093
$endgroup$
$begingroup$
Thank you, this conversion never occurred to me
$endgroup$
– sdds
Jan 30 at 16:37
$begingroup$
@sdds You're welcome!
$endgroup$
– Clement C.
Jan 30 at 16:40
add a comment |
$begingroup$
$$frac{ln(1+2x)sin{x}}{sqrt{x^3}}=frac{ln(1+2x)}{2x}cdotfrac{sin{x}}{x}cdot2sqrt{x}rightarrow0.$$
answered Jan 30 at 16:24
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
It is known that
$$lim_{xto 0^+}frac{sin x}x=1$$ and $$lim_{xto 0^+}frac{log(1+x)}x=1=lim_{xto 0^+}frac{log(1+2x)}{2x}.$$
Then
$$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}}=1cdot2cdotlim_{xto 0^+} sqrt x.$$
answered Jan 30 at 16:42
Yves DaoustYves Daoust
132k676229
$endgroup$
2
$begingroup$
It's true, but how is that different from the current two answers?
$endgroup$
– Clement C.
Jan 30 at 16:45
add a comment |
$begingroup$
If you want to, you should apply L'Hospital's rule twice:
$$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}}=\
lim_{xto 0^+} frac{frac{2}{1+2x}sin x+ln(1+2x)cos x}{1.5sqrt {x}}=\
lim_{xto 0^+} frac{frac{-4}{(1+2x)^2}sin x+frac{2}{1+2x}cos x+frac{2}{1+2x}cos x-ln(1+2x)sin x}{frac{0.75}{sqrt {x}}}=0.
$$
answered Feb 28 at 10:15
farruhotafarruhota
21.8k2842
$endgroup$
add a comment |
Your Answer
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Post as a guest
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Use the facts that (1) $lim_{xto 0} frac{ln(1+2x)}{2x} = 1$, (2) $lim_{xto 0} frac{sin x}{x} = 1$, and (3) $lim_{xto 0^+} frac{x^2}{sqrt{x^3}} = 0$ to show the limit is $2cdot 1cdot 0 = 0$.
To do so, rewrite, for $x>0$,
$$
frac{ln(1+2x)sin x}{sqrt{x^3}}=2cdot frac{ln(1+2x)}{2x}cdotfrac{sin x}{x}cdotfrac{x^2}{sqrt{x^3}}
$$
answered Jan 30 at 16:24
Clement C.Clement C.
51k34093
$endgroup$
$begingroup$
Thank you, this conversion never occurred to me
$endgroup$
– sdds
Jan 30 at 16:37
$begingroup$
@sdds You're welcome!
$endgroup$
– Clement C.
Jan 30 at 16:40
add a comment |
$begingroup$
Hint:
Use the facts that (1) $lim_{xto 0} frac{ln(1+2x)}{2x} = 1$, (2) $lim_{xto 0} frac{sin x}{x} = 1$, and (3) $lim_{xto 0^+} frac{x^2}{sqrt{x^3}} = 0$ to show the limit is $2cdot 1cdot 0 = 0$.
To do so, rewrite, for $x>0$,
$$
frac{ln(1+2x)sin x}{sqrt{x^3}}=2cdot frac{ln(1+2x)}{2x}cdotfrac{sin x}{x}cdotfrac{x^2}{sqrt{x^3}}
$$
answered Jan 30 at 16:24
Clement C.Clement C.
51k34093
$endgroup$
$begingroup$
Thank you, this conversion never occurred to me
$endgroup$
– sdds
Jan 30 at 16:37
$begingroup$
@sdds You're welcome!
$endgroup$
– Clement C.
Jan 30 at 16:40
add a comment |
$begingroup$
Hint:
Use the facts that (1) $lim_{xto 0} frac{ln(1+2x)}{2x} = 1$, (2) $lim_{xto 0} frac{sin x}{x} = 1$, and (3) $lim_{xto 0^+} frac{x^2}{sqrt{x^3}} = 0$ to show the limit is $2cdot 1cdot 0 = 0$.
To do so, rewrite, for $x>0$,
$$
frac{ln(1+2x)sin x}{sqrt{x^3}}=2cdot frac{ln(1+2x)}{2x}cdotfrac{sin x}{x}cdotfrac{x^2}{sqrt{x^3}}
$$
answered Jan 30 at 16:24
Clement C.Clement C.
51k34093
$endgroup$
Hint:
Use the facts that (1) $lim_{xto 0} frac{ln(1+2x)}{2x} = 1$, (2) $lim_{xto 0} frac{sin x}{x} = 1$, and (3) $lim_{xto 0^+} frac{x^2}{sqrt{x^3}} = 0$ to show the limit is $2cdot 1cdot 0 = 0$.
To do so, rewrite, for $x>0$,
$$
frac{ln(1+2x)sin x}{sqrt{x^3}}=2cdot frac{ln(1+2x)}{2x}cdotfrac{sin x}{x}cdotfrac{x^2}{sqrt{x^3}}
$$
answered Jan 30 at 16:24
Clement C.Clement C.
51k34093
answered Jan 30 at 16:24
Clement C.Clement C.
51k34093
answered Jan 30 at 16:24
Clement C.Clement C.
51k34093
answered Jan 30 at 16:24
Clement C.Clement C.
51k34093
51k34093
$begingroup$
Thank you, this conversion never occurred to me
$endgroup$
– sdds
Jan 30 at 16:37
$begingroup$
@sdds You're welcome!
$endgroup$
– Clement C.
Jan 30 at 16:40
add a comment |
$begingroup$
Thank you, this conversion never occurred to me
$endgroup$
– sdds
Jan 30 at 16:37
$begingroup$
@sdds You're welcome!
$endgroup$
– Clement C.
Jan 30 at 16:40
$begingroup$
Thank you, this conversion never occurred to me
$endgroup$
– sdds
Jan 30 at 16:37
$begingroup$
Thank you, this conversion never occurred to me
$endgroup$
– sdds
Jan 30 at 16:37
$begingroup$
@sdds You're welcome!
$endgroup$
– Clement C.
Jan 30 at 16:40
$begingroup$
@sdds You're welcome!
$endgroup$
– Clement C.
Jan 30 at 16:40
add a comment |
$begingroup$
$$frac{ln(1+2x)sin{x}}{sqrt{x^3}}=frac{ln(1+2x)}{2x}cdotfrac{sin{x}}{x}cdot2sqrt{x}rightarrow0.$$
answered Jan 30 at 16:24
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
$$frac{ln(1+2x)sin{x}}{sqrt{x^3}}=frac{ln(1+2x)}{2x}cdotfrac{sin{x}}{x}cdot2sqrt{x}rightarrow0.$$
answered Jan 30 at 16:24
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
$$frac{ln(1+2x)sin{x}}{sqrt{x^3}}=frac{ln(1+2x)}{2x}cdotfrac{sin{x}}{x}cdot2sqrt{x}rightarrow0.$$
answered Jan 30 at 16:24
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
$$frac{ln(1+2x)sin{x}}{sqrt{x^3}}=frac{ln(1+2x)}{2x}cdotfrac{sin{x}}{x}cdot2sqrt{x}rightarrow0.$$
answered Jan 30 at 16:24
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 30 at 16:24
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 30 at 16:24
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 30 at 16:24
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
add a comment |
add a comment |
$begingroup$
It is known that
$$lim_{xto 0^+}frac{sin x}x=1$$ and $$lim_{xto 0^+}frac{log(1+x)}x=1=lim_{xto 0^+}frac{log(1+2x)}{2x}.$$
Then
$$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}}=1cdot2cdotlim_{xto 0^+} sqrt x.$$
answered Jan 30 at 16:42
Yves DaoustYves Daoust
132k676229
$endgroup$
2
$begingroup$
It's true, but how is that different from the current two answers?
$endgroup$
– Clement C.
Jan 30 at 16:45
add a comment |
$begingroup$
It is known that
$$lim_{xto 0^+}frac{sin x}x=1$$ and $$lim_{xto 0^+}frac{log(1+x)}x=1=lim_{xto 0^+}frac{log(1+2x)}{2x}.$$
Then
$$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}}=1cdot2cdotlim_{xto 0^+} sqrt x.$$
answered Jan 30 at 16:42
Yves DaoustYves Daoust
132k676229
$endgroup$
2
$begingroup$
It's true, but how is that different from the current two answers?
$endgroup$
– Clement C.
Jan 30 at 16:45
add a comment |
$begingroup$
It is known that
$$lim_{xto 0^+}frac{sin x}x=1$$ and $$lim_{xto 0^+}frac{log(1+x)}x=1=lim_{xto 0^+}frac{log(1+2x)}{2x}.$$
Then
$$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}}=1cdot2cdotlim_{xto 0^+} sqrt x.$$
answered Jan 30 at 16:42
Yves DaoustYves Daoust
132k676229
$endgroup$
It is known that
$$lim_{xto 0^+}frac{sin x}x=1$$ and $$lim_{xto 0^+}frac{log(1+x)}x=1=lim_{xto 0^+}frac{log(1+2x)}{2x}.$$
Then
$$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}}=1cdot2cdotlim_{xto 0^+} sqrt x.$$
answered Jan 30 at 16:42
Yves DaoustYves Daoust
132k676229
answered Jan 30 at 16:42
Yves DaoustYves Daoust
132k676229
answered Jan 30 at 16:42
Yves DaoustYves Daoust
132k676229
answered Jan 30 at 16:42
Yves DaoustYves Daoust
132k676229
132k676229
2
$begingroup$
It's true, but how is that different from the current two answers?
$endgroup$
– Clement C.
Jan 30 at 16:45
add a comment |
2
$begingroup$
It's true, but how is that different from the current two answers?
$endgroup$
– Clement C.
Jan 30 at 16:45
2
2
$begingroup$
It's true, but how is that different from the current two answers?
$endgroup$
– Clement C.
Jan 30 at 16:45
$begingroup$
It's true, but how is that different from the current two answers?
$endgroup$
– Clement C.
Jan 30 at 16:45
add a comment |
$begingroup$
If you want to, you should apply L'Hospital's rule twice:
$$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}}=\
lim_{xto 0^+} frac{frac{2}{1+2x}sin x+ln(1+2x)cos x}{1.5sqrt {x}}=\
lim_{xto 0^+} frac{frac{-4}{(1+2x)^2}sin x+frac{2}{1+2x}cos x+frac{2}{1+2x}cos x-ln(1+2x)sin x}{frac{0.75}{sqrt {x}}}=0.
$$
answered Feb 28 at 10:15
farruhotafarruhota
21.8k2842
$endgroup$
add a comment |
$begingroup$
If you want to, you should apply L'Hospital's rule twice:
$$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}}=\
lim_{xto 0^+} frac{frac{2}{1+2x}sin x+ln(1+2x)cos x}{1.5sqrt {x}}=\
lim_{xto 0^+} frac{frac{-4}{(1+2x)^2}sin x+frac{2}{1+2x}cos x+frac{2}{1+2x}cos x-ln(1+2x)sin x}{frac{0.75}{sqrt {x}}}=0.
$$
answered Feb 28 at 10:15
farruhotafarruhota
21.8k2842
$endgroup$
add a comment |
$begingroup$
If you want to, you should apply L'Hospital's rule twice:
$$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}}=\
lim_{xto 0^+} frac{frac{2}{1+2x}sin x+ln(1+2x)cos x}{1.5sqrt {x}}=\
lim_{xto 0^+} frac{frac{-4}{(1+2x)^2}sin x+frac{2}{1+2x}cos x+frac{2}{1+2x}cos x-ln(1+2x)sin x}{frac{0.75}{sqrt {x}}}=0.
$$
answered Feb 28 at 10:15
farruhotafarruhota
21.8k2842
$endgroup$
If you want to, you should apply L'Hospital's rule twice:
$$lim_{xto 0^+} frac{ln(1+2x)sin x}{sqrt {x^3}}=\
lim_{xto 0^+} frac{frac{2}{1+2x}sin x+ln(1+2x)cos x}{1.5sqrt {x}}=\
lim_{xto 0^+} frac{frac{-4}{(1+2x)^2}sin x+frac{2}{1+2x}cos x+frac{2}{1+2x}cos x-ln(1+2x)sin x}{frac{0.75}{sqrt {x}}}=0.
$$
answered Feb 28 at 10:15
farruhotafarruhota
21.8k2842
answered Feb 28 at 10:15
farruhotafarruhota
21.8k2842
answered Feb 28 at 10:15
farruhotafarruhota
21.8k2842
answered Feb 28 at 10:15
farruhotafarruhota
21.8k2842
21.8k2842
add a comment |
add a comment |
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Using Taylor, the numerator is $2xcdot x+o(x^3)$ and the limit is $0$.
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– Yves Daoust
Jan 30 at 16:37
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We haven't been taught about Taylor yet, I should add this to my question. However, thanks for the answer, i'll research Taylor on my own.
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– sdds
Jan 30 at 16:40
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To answer the question in the title, see for example here
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– BlueRaja - Danny Pflughoeft
Jan 30 at 21:37
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After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
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– Clement C.
Feb 3 at 23:26