Help understanding the steps of a solved limit
A friend of mine gave me this already solved limit and I'm trying to understand all the steps that he did to solve it, here's the limit:
begin{align}
lim_{x rightarrow +∞}
frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-sqrt{1/x}}
&=lim_{x rightarrow +∞}
frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
\
&=
lim_{x rightarrow +∞}
frac{2x}{(6x^3)}= 0
end{align}
The part that I don't understand is why the limit is equal to:
begin{equation*}
frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
end{equation*}
The only thing I found out so far is that:
begin{equation*}
sin (x) - x ∼ x^3/6
end{equation*}
So:
begin{equation*}
sin (1/x) - (1/x)∼ 1/(6x^3)
end{equation*}
For the rest, I have no idea .
limits asymptotics
asked Dec 22 '18 at 14:08
El Bryan
397
add a comment |
A friend of mine gave me this already solved limit and I'm trying to understand all the steps that he did to solve it, here's the limit:
begin{align}
lim_{x rightarrow +∞}
frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-sqrt{1/x}}
&=lim_{x rightarrow +∞}
frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
\
&=
lim_{x rightarrow +∞}
frac{2x}{(6x^3)}= 0
end{align}
The part that I don't understand is why the limit is equal to:
begin{equation*}
frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
end{equation*}
The only thing I found out so far is that:
begin{equation*}
sin (x) - x ∼ x^3/6
end{equation*}
So:
begin{equation*}
sin (1/x) - (1/x)∼ 1/(6x^3)
end{equation*}
For the rest, I have no idea .
limits asymptotics
asked Dec 22 '18 at 14:08
El Bryan
397
add a comment |
A friend of mine gave me this already solved limit and I'm trying to understand all the steps that he did to solve it, here's the limit:
begin{align}
lim_{x rightarrow +∞}
frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-sqrt{1/x}}
&=lim_{x rightarrow +∞}
frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
\
&=
lim_{x rightarrow +∞}
frac{2x}{(6x^3)}= 0
end{align}
The part that I don't understand is why the limit is equal to:
begin{equation*}
frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
end{equation*}
The only thing I found out so far is that:
begin{equation*}
sin (x) - x ∼ x^3/6
end{equation*}
So:
begin{equation*}
sin (1/x) - (1/x)∼ 1/(6x^3)
end{equation*}
For the rest, I have no idea .
limits asymptotics
asked Dec 22 '18 at 14:08
El Bryan
397
A friend of mine gave me this already solved limit and I'm trying to understand all the steps that he did to solve it, here's the limit:
begin{align}
lim_{x rightarrow +∞}
frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-sqrt{1/x}}
&=lim_{x rightarrow +∞}
frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
\
&=
lim_{x rightarrow +∞}
frac{2x}{(6x^3)}= 0
end{align}
The part that I don't understand is why the limit is equal to:
begin{equation*}
frac{(1/x) - (1/(6x^3))-(1/x)}{(1/x)-(1/(2x))-(1/x)}
end{equation*}
The only thing I found out so far is that:
begin{equation*}
sin (x) - x ∼ x^3/6
end{equation*}
So:
begin{equation*}
sin (1/x) - (1/x)∼ 1/(6x^3)
end{equation*}
For the rest, I have no idea .
limits asymptotics
limits asymptotics
asked Dec 22 '18 at 14:08
El Bryan
397
asked Dec 22 '18 at 14:08
El Bryan
397
edited Dec 22 '18 at 15:09
asked Dec 22 '18 at 14:08
El Bryan
397
asked Dec 22 '18 at 14:08
El Bryan
397
asked Dec 22 '18 at 14:08
El Bryan
397
397
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
It isn't exactly correct.
$$sin(x)=x-x^3/3!+x^5/5!-cdots$$
This can be obtained from Taylor's expansion.
For $log(1+x),$
$$log(1+x)=intfrac{1}{1+x}dx$$
$$=int (1-x+x^2-x^3+cdots)dx$$
$$log(1+x)=x-x^2/2+cdots$$
So, your limit,
begin{equation*}
lim_{x rightarrow +∞}
frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-(1/sqrt x)}=
end{equation*}
begin{equation*}
lim_{x rightarrow +∞}
frac{(1/x) -(1/6x^3)cdots - (1/x)}{(1/sqrt x)-(1/2x)cdots-(1/sqrt x)}=
end{equation*}
Since the limit is to infinity, only the coeff. of highest powers (of -3 and -1 in this case) matter.
begin{equation*}
lim_{x rightarrow +∞}
frac{2x}{6x^3}= 0
end{equation*}
Your solution has a lot of typos
edited Dec 22 '18 at 14:54
Bernard
118k639112
answered Dec 22 '18 at 14:30
Ankit Kumar
1
You used taylor? also what are typos (I'm new here)
– El Bryan
Dec 22 '18 at 14:42
I've editted my answer to include what you asked. You've 2 typos. First, in last term of denominator of very first limit. Second, you've written $6/x^3$ instead of $6x^3$ in the 3rd limit
– Ankit Kumar
Dec 22 '18 at 14:47
1
Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
– El Bryan
Dec 22 '18 at 15:11
@ElBryan I'm not sure that you can use asymptotic equivalence in this question. It's useful for cases like $x^2+x$, which you can approx. as $x^2$, as x goes to infinity.
– Ankit Kumar
Dec 22 '18 at 15:15
add a comment |
First note that $xto infty$, perform substitution $t = {1over x}$, then $tto 0$:
$$
lim_{tto 0} frac{sin t - t}{log(1 + sqrt{t}) - sqrt{t}}
$$
By Taylor expansion you may approximate $sin t$ at $t = 0$ by:
$$
sin t = t - {t^3over 3!} + {t^5over 5!} - dots
$$
At the same time for $log(1+t)$:
$$
log(1+t) = t - {t^2over 2} + {t^3over 3}-cdots
$$
So if you apply this to your limit you'll observe the desired result:
$$
begin{align}
lim_{tto 0} frac{t - {t^3over 3!} - t}{sqrt{t} - {tover 2} - sqrt{t}} &= lim_{tto 0} frac{{t^3over 3!}}{{tover 2}} = \
&= lim_{tto 0} frac{2t^3}{6t} = \
&= lim_{tto 0} frac{2t^2}{6} = 0
end{align}
$$
answered Dec 22 '18 at 14:29
roman
1,98121221
add a comment |
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2 Answers
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2 Answers
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oldest
votes
It isn't exactly correct.
$$sin(x)=x-x^3/3!+x^5/5!-cdots$$
This can be obtained from Taylor's expansion.
For $log(1+x),$
$$log(1+x)=intfrac{1}{1+x}dx$$
$$=int (1-x+x^2-x^3+cdots)dx$$
$$log(1+x)=x-x^2/2+cdots$$
So, your limit,
begin{equation*}
lim_{x rightarrow +∞}
frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-(1/sqrt x)}=
end{equation*}
begin{equation*}
lim_{x rightarrow +∞}
frac{(1/x) -(1/6x^3)cdots - (1/x)}{(1/sqrt x)-(1/2x)cdots-(1/sqrt x)}=
end{equation*}
Since the limit is to infinity, only the coeff. of highest powers (of -3 and -1 in this case) matter.
begin{equation*}
lim_{x rightarrow +∞}
frac{2x}{6x^3}= 0
end{equation*}
Your solution has a lot of typos
edited Dec 22 '18 at 14:54
Bernard
118k639112
answered Dec 22 '18 at 14:30
Ankit Kumar
1
You used taylor? also what are typos (I'm new here)
– El Bryan
Dec 22 '18 at 14:42
I've editted my answer to include what you asked. You've 2 typos. First, in last term of denominator of very first limit. Second, you've written $6/x^3$ instead of $6x^3$ in the 3rd limit
– Ankit Kumar
Dec 22 '18 at 14:47
1
Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
– El Bryan
Dec 22 '18 at 15:11
@ElBryan I'm not sure that you can use asymptotic equivalence in this question. It's useful for cases like $x^2+x$, which you can approx. as $x^2$, as x goes to infinity.
– Ankit Kumar
Dec 22 '18 at 15:15
add a comment |
It isn't exactly correct.
$$sin(x)=x-x^3/3!+x^5/5!-cdots$$
This can be obtained from Taylor's expansion.
For $log(1+x),$
$$log(1+x)=intfrac{1}{1+x}dx$$
$$=int (1-x+x^2-x^3+cdots)dx$$
$$log(1+x)=x-x^2/2+cdots$$
So, your limit,
begin{equation*}
lim_{x rightarrow +∞}
frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-(1/sqrt x)}=
end{equation*}
begin{equation*}
lim_{x rightarrow +∞}
frac{(1/x) -(1/6x^3)cdots - (1/x)}{(1/sqrt x)-(1/2x)cdots-(1/sqrt x)}=
end{equation*}
Since the limit is to infinity, only the coeff. of highest powers (of -3 and -1 in this case) matter.
begin{equation*}
lim_{x rightarrow +∞}
frac{2x}{6x^3}= 0
end{equation*}
Your solution has a lot of typos
edited Dec 22 '18 at 14:54
Bernard
118k639112
answered Dec 22 '18 at 14:30
Ankit Kumar
1
You used taylor? also what are typos (I'm new here)
– El Bryan
Dec 22 '18 at 14:42
I've editted my answer to include what you asked. You've 2 typos. First, in last term of denominator of very first limit. Second, you've written $6/x^3$ instead of $6x^3$ in the 3rd limit
– Ankit Kumar
Dec 22 '18 at 14:47
1
Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
– El Bryan
Dec 22 '18 at 15:11
@ElBryan I'm not sure that you can use asymptotic equivalence in this question. It's useful for cases like $x^2+x$, which you can approx. as $x^2$, as x goes to infinity.
– Ankit Kumar
Dec 22 '18 at 15:15
add a comment |
It isn't exactly correct.
$$sin(x)=x-x^3/3!+x^5/5!-cdots$$
This can be obtained from Taylor's expansion.
For $log(1+x),$
$$log(1+x)=intfrac{1}{1+x}dx$$
$$=int (1-x+x^2-x^3+cdots)dx$$
$$log(1+x)=x-x^2/2+cdots$$
So, your limit,
begin{equation*}
lim_{x rightarrow +∞}
frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-(1/sqrt x)}=
end{equation*}
begin{equation*}
lim_{x rightarrow +∞}
frac{(1/x) -(1/6x^3)cdots - (1/x)}{(1/sqrt x)-(1/2x)cdots-(1/sqrt x)}=
end{equation*}
Since the limit is to infinity, only the coeff. of highest powers (of -3 and -1 in this case) matter.
begin{equation*}
lim_{x rightarrow +∞}
frac{2x}{6x^3}= 0
end{equation*}
Your solution has a lot of typos
edited Dec 22 '18 at 14:54
Bernard
118k639112
answered Dec 22 '18 at 14:30
Ankit Kumar
1
It isn't exactly correct.
$$sin(x)=x-x^3/3!+x^5/5!-cdots$$
This can be obtained from Taylor's expansion.
For $log(1+x),$
$$log(1+x)=intfrac{1}{1+x}dx$$
$$=int (1-x+x^2-x^3+cdots)dx$$
$$log(1+x)=x-x^2/2+cdots$$
So, your limit,
begin{equation*}
lim_{x rightarrow +∞}
frac{sin (1/x) - (1/x)}{log(1+(1/sqrt{x}))-(1/sqrt x)}=
end{equation*}
begin{equation*}
lim_{x rightarrow +∞}
frac{(1/x) -(1/6x^3)cdots - (1/x)}{(1/sqrt x)-(1/2x)cdots-(1/sqrt x)}=
end{equation*}
Since the limit is to infinity, only the coeff. of highest powers (of -3 and -1 in this case) matter.
begin{equation*}
lim_{x rightarrow +∞}
frac{2x}{6x^3}= 0
end{equation*}
Your solution has a lot of typos
edited Dec 22 '18 at 14:54
Bernard
118k639112
answered Dec 22 '18 at 14:30
Ankit Kumar
1
edited Dec 22 '18 at 14:54
Bernard
118k639112
edited Dec 22 '18 at 14:54
Bernard
118k639112
edited Dec 22 '18 at 14:54
Bernard
118k639112
118k639112
answered Dec 22 '18 at 14:30
Ankit Kumar
1
answered Dec 22 '18 at 14:30
Ankit Kumar
1
answered Dec 22 '18 at 14:30
Ankit Kumar
1
1
You used taylor? also what are typos (I'm new here)
– El Bryan
Dec 22 '18 at 14:42
I've editted my answer to include what you asked. You've 2 typos. First, in last term of denominator of very first limit. Second, you've written $6/x^3$ instead of $6x^3$ in the 3rd limit
– Ankit Kumar
Dec 22 '18 at 14:47
1
Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
– El Bryan
Dec 22 '18 at 15:11
@ElBryan I'm not sure that you can use asymptotic equivalence in this question. It's useful for cases like $x^2+x$, which you can approx. as $x^2$, as x goes to infinity.
– Ankit Kumar
Dec 22 '18 at 15:15
add a comment |
You used taylor? also what are typos (I'm new here)
– El Bryan
Dec 22 '18 at 14:42
I've editted my answer to include what you asked. You've 2 typos. First, in last term of denominator of very first limit. Second, you've written $6/x^3$ instead of $6x^3$ in the 3rd limit
– Ankit Kumar
Dec 22 '18 at 14:47
1
Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
– El Bryan
Dec 22 '18 at 15:11
@ElBryan I'm not sure that you can use asymptotic equivalence in this question. It's useful for cases like $x^2+x$, which you can approx. as $x^2$, as x goes to infinity.
– Ankit Kumar
Dec 22 '18 at 15:15
You used taylor? also what are typos (I'm new here)
– El Bryan
Dec 22 '18 at 14:42
You used taylor? also what are typos (I'm new here)
– El Bryan
Dec 22 '18 at 14:42
I've editted my answer to include what you asked. You've 2 typos. First, in last term of denominator of very first limit. Second, you've written $6/x^3$ instead of $6x^3$ in the 3rd limit
– Ankit Kumar
Dec 22 '18 at 14:47
I've editted my answer to include what you asked. You've 2 typos. First, in last term of denominator of very first limit. Second, you've written $6/x^3$ instead of $6x^3$ in the 3rd limit
– Ankit Kumar
Dec 22 '18 at 14:47
1
1
Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
– El Bryan
Dec 22 '18 at 15:11
Ok, this is the answer I was looking for, one more thing , is it possible to solve it with asymptotic equivalence?
– El Bryan
Dec 22 '18 at 15:11
@ElBryan I'm not sure that you can use asymptotic equivalence in this question. It's useful for cases like $x^2+x$, which you can approx. as $x^2$, as x goes to infinity.
– Ankit Kumar
Dec 22 '18 at 15:15
@ElBryan I'm not sure that you can use asymptotic equivalence in this question. It's useful for cases like $x^2+x$, which you can approx. as $x^2$, as x goes to infinity.
– Ankit Kumar
Dec 22 '18 at 15:15
add a comment |
First note that $xto infty$, perform substitution $t = {1over x}$, then $tto 0$:
$$
lim_{tto 0} frac{sin t - t}{log(1 + sqrt{t}) - sqrt{t}}
$$
By Taylor expansion you may approximate $sin t$ at $t = 0$ by:
$$
sin t = t - {t^3over 3!} + {t^5over 5!} - dots
$$
At the same time for $log(1+t)$:
$$
log(1+t) = t - {t^2over 2} + {t^3over 3}-cdots
$$
So if you apply this to your limit you'll observe the desired result:
$$
begin{align}
lim_{tto 0} frac{t - {t^3over 3!} - t}{sqrt{t} - {tover 2} - sqrt{t}} &= lim_{tto 0} frac{{t^3over 3!}}{{tover 2}} = \
&= lim_{tto 0} frac{2t^3}{6t} = \
&= lim_{tto 0} frac{2t^2}{6} = 0
end{align}
$$
answered Dec 22 '18 at 14:29
roman
1,98121221
add a comment |
First note that $xto infty$, perform substitution $t = {1over x}$, then $tto 0$:
$$
lim_{tto 0} frac{sin t - t}{log(1 + sqrt{t}) - sqrt{t}}
$$
By Taylor expansion you may approximate $sin t$ at $t = 0$ by:
$$
sin t = t - {t^3over 3!} + {t^5over 5!} - dots
$$
At the same time for $log(1+t)$:
$$
log(1+t) = t - {t^2over 2} + {t^3over 3}-cdots
$$
So if you apply this to your limit you'll observe the desired result:
$$
begin{align}
lim_{tto 0} frac{t - {t^3over 3!} - t}{sqrt{t} - {tover 2} - sqrt{t}} &= lim_{tto 0} frac{{t^3over 3!}}{{tover 2}} = \
&= lim_{tto 0} frac{2t^3}{6t} = \
&= lim_{tto 0} frac{2t^2}{6} = 0
end{align}
$$
answered Dec 22 '18 at 14:29
roman
1,98121221
add a comment |
First note that $xto infty$, perform substitution $t = {1over x}$, then $tto 0$:
$$
lim_{tto 0} frac{sin t - t}{log(1 + sqrt{t}) - sqrt{t}}
$$
By Taylor expansion you may approximate $sin t$ at $t = 0$ by:
$$
sin t = t - {t^3over 3!} + {t^5over 5!} - dots
$$
At the same time for $log(1+t)$:
$$
log(1+t) = t - {t^2over 2} + {t^3over 3}-cdots
$$
So if you apply this to your limit you'll observe the desired result:
$$
begin{align}
lim_{tto 0} frac{t - {t^3over 3!} - t}{sqrt{t} - {tover 2} - sqrt{t}} &= lim_{tto 0} frac{{t^3over 3!}}{{tover 2}} = \
&= lim_{tto 0} frac{2t^3}{6t} = \
&= lim_{tto 0} frac{2t^2}{6} = 0
end{align}
$$
answered Dec 22 '18 at 14:29
roman
1,98121221
First note that $xto infty$, perform substitution $t = {1over x}$, then $tto 0$:
$$
lim_{tto 0} frac{sin t - t}{log(1 + sqrt{t}) - sqrt{t}}
$$
By Taylor expansion you may approximate $sin t$ at $t = 0$ by:
$$
sin t = t - {t^3over 3!} + {t^5over 5!} - dots
$$
At the same time for $log(1+t)$:
$$
log(1+t) = t - {t^2over 2} + {t^3over 3}-cdots
$$
So if you apply this to your limit you'll observe the desired result:
$$
begin{align}
lim_{tto 0} frac{t - {t^3over 3!} - t}{sqrt{t} - {tover 2} - sqrt{t}} &= lim_{tto 0} frac{{t^3over 3!}}{{tover 2}} = \
&= lim_{tto 0} frac{2t^3}{6t} = \
&= lim_{tto 0} frac{2t^2}{6} = 0
end{align}
$$
answered Dec 22 '18 at 14:29
roman
1,98121221
edited Dec 22 '18 at 14:55
answered Dec 22 '18 at 14:29
roman
1,98121221
answered Dec 22 '18 at 14:29
roman
1,98121221
answered Dec 22 '18 at 14:29
roman
1,98121221
1,98121221
add a comment |
add a comment |
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