Need a hint evaluating $ limlimits_{xto 0}frac{xln{(frac{sin (x)}{x})}}{sin (x) - x} $
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1
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I'm stuck with this. I've tried substituting $t$ for $frac{sin (x)}{x}$ and $sin (x) - x$ but it doesn't work at all.
A small hint would be greatly appreciated.
limits limits-without-lhopital
asked Nov 21 at 23:29
Bartosz
315
add a comment |
up vote
1
down vote
favorite
I'm stuck with this. I've tried substituting $t$ for $frac{sin (x)}{x}$ and $sin (x) - x$ but it doesn't work at all.
A small hint would be greatly appreciated.
limits limits-without-lhopital
asked Nov 21 at 23:29
Bartosz
315
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm stuck with this. I've tried substituting $t$ for $frac{sin (x)}{x}$ and $sin (x) - x$ but it doesn't work at all.
A small hint would be greatly appreciated.
limits limits-without-lhopital
asked Nov 21 at 23:29
Bartosz
315
I'm stuck with this. I've tried substituting $t$ for $frac{sin (x)}{x}$ and $sin (x) - x$ but it doesn't work at all.
A small hint would be greatly appreciated.
limits limits-without-lhopital
limits limits-without-lhopital
asked Nov 21 at 23:29
Bartosz
315
asked Nov 21 at 23:29
Bartosz
315
asked Nov 21 at 23:29
Bartosz
315
asked Nov 21 at 23:29
Bartosz
315
asked Nov 21 at 23:29
Bartosz
315
315
add a comment |
add a comment |
2 Answers
2
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oldest
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up vote
2
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accepted
$$lim_{xto 0}frac{xln{(frac{sin (x)}{x})}}{sin (x) - x} = lim_{xto 0}frac{ln{(1+frac{sin (x)-x}{x})}}{frac{sin (x) - x}{x}} =1 $$ Because $$lim_{xto0}frac{sin x-x}{x}=lim_{xto0}frac{sin x}{x}-1=1-1=0$$ and $$lim_{tto0}frac{ln(1+t)}{t}=1.$$
answered Nov 21 at 23:34
Tito Eliatron
1,317622
1
Woah, it's so elegant. Thank you very much
– Bartosz
Nov 21 at 23:36
add a comment |
up vote
0
down vote
Use the Taylor expansion
$$
frac{sin x}{x}=1-frac{x^2}{6}+o(x^2)
$$
to conclude that
$$
xlnfrac{sin x}{x}=-frac{x^3}{6}+o(x^3)
$$
Likewise
$$
sin x-x=x-frac{x^3}{6}+o(x^3)-x=-frac{x^3}{6}+o(x^3)
$$
So the limit is $1$.
answered Nov 21 at 23:41
egreg
176k1384198
We didn't cover Taylor expansion (not even derivatives) yet, but thank you for contribution :)
– Bartosz
Nov 21 at 23:49
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$$lim_{xto 0}frac{xln{(frac{sin (x)}{x})}}{sin (x) - x} = lim_{xto 0}frac{ln{(1+frac{sin (x)-x}{x})}}{frac{sin (x) - x}{x}} =1 $$ Because $$lim_{xto0}frac{sin x-x}{x}=lim_{xto0}frac{sin x}{x}-1=1-1=0$$ and $$lim_{tto0}frac{ln(1+t)}{t}=1.$$
answered Nov 21 at 23:34
Tito Eliatron
1,317622
1
Woah, it's so elegant. Thank you very much
– Bartosz
Nov 21 at 23:36
add a comment |
up vote
2
down vote
accepted
$$lim_{xto 0}frac{xln{(frac{sin (x)}{x})}}{sin (x) - x} = lim_{xto 0}frac{ln{(1+frac{sin (x)-x}{x})}}{frac{sin (x) - x}{x}} =1 $$ Because $$lim_{xto0}frac{sin x-x}{x}=lim_{xto0}frac{sin x}{x}-1=1-1=0$$ and $$lim_{tto0}frac{ln(1+t)}{t}=1.$$
answered Nov 21 at 23:34
Tito Eliatron
1,317622
1
Woah, it's so elegant. Thank you very much
– Bartosz
Nov 21 at 23:36
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$$lim_{xto 0}frac{xln{(frac{sin (x)}{x})}}{sin (x) - x} = lim_{xto 0}frac{ln{(1+frac{sin (x)-x}{x})}}{frac{sin (x) - x}{x}} =1 $$ Because $$lim_{xto0}frac{sin x-x}{x}=lim_{xto0}frac{sin x}{x}-1=1-1=0$$ and $$lim_{tto0}frac{ln(1+t)}{t}=1.$$
answered Nov 21 at 23:34
Tito Eliatron
1,317622
$$lim_{xto 0}frac{xln{(frac{sin (x)}{x})}}{sin (x) - x} = lim_{xto 0}frac{ln{(1+frac{sin (x)-x}{x})}}{frac{sin (x) - x}{x}} =1 $$ Because $$lim_{xto0}frac{sin x-x}{x}=lim_{xto0}frac{sin x}{x}-1=1-1=0$$ and $$lim_{tto0}frac{ln(1+t)}{t}=1.$$
answered Nov 21 at 23:34
Tito Eliatron
1,317622
answered Nov 21 at 23:34
Tito Eliatron
1,317622
answered Nov 21 at 23:34
Tito Eliatron
1,317622
answered Nov 21 at 23:34
Tito Eliatron
1,317622
1,317622
1
Woah, it's so elegant. Thank you very much
– Bartosz
Nov 21 at 23:36
add a comment |
1
Woah, it's so elegant. Thank you very much
– Bartosz
Nov 21 at 23:36
1
1
Woah, it's so elegant. Thank you very much
– Bartosz
Nov 21 at 23:36
Woah, it's so elegant. Thank you very much
– Bartosz
Nov 21 at 23:36
add a comment |
up vote
0
down vote
Use the Taylor expansion
$$
frac{sin x}{x}=1-frac{x^2}{6}+o(x^2)
$$
to conclude that
$$
xlnfrac{sin x}{x}=-frac{x^3}{6}+o(x^3)
$$
Likewise
$$
sin x-x=x-frac{x^3}{6}+o(x^3)-x=-frac{x^3}{6}+o(x^3)
$$
So the limit is $1$.
answered Nov 21 at 23:41
egreg
176k1384198
We didn't cover Taylor expansion (not even derivatives) yet, but thank you for contribution :)
– Bartosz
Nov 21 at 23:49
add a comment |
up vote
0
down vote
Use the Taylor expansion
$$
frac{sin x}{x}=1-frac{x^2}{6}+o(x^2)
$$
to conclude that
$$
xlnfrac{sin x}{x}=-frac{x^3}{6}+o(x^3)
$$
Likewise
$$
sin x-x=x-frac{x^3}{6}+o(x^3)-x=-frac{x^3}{6}+o(x^3)
$$
So the limit is $1$.
answered Nov 21 at 23:41
egreg
176k1384198
We didn't cover Taylor expansion (not even derivatives) yet, but thank you for contribution :)
– Bartosz
Nov 21 at 23:49
add a comment |
up vote
0
down vote
up vote
0
down vote
Use the Taylor expansion
$$
frac{sin x}{x}=1-frac{x^2}{6}+o(x^2)
$$
to conclude that
$$
xlnfrac{sin x}{x}=-frac{x^3}{6}+o(x^3)
$$
Likewise
$$
sin x-x=x-frac{x^3}{6}+o(x^3)-x=-frac{x^3}{6}+o(x^3)
$$
So the limit is $1$.
answered Nov 21 at 23:41
egreg
176k1384198
Use the Taylor expansion
$$
frac{sin x}{x}=1-frac{x^2}{6}+o(x^2)
$$
to conclude that
$$
xlnfrac{sin x}{x}=-frac{x^3}{6}+o(x^3)
$$
Likewise
$$
sin x-x=x-frac{x^3}{6}+o(x^3)-x=-frac{x^3}{6}+o(x^3)
$$
So the limit is $1$.
answered Nov 21 at 23:41
egreg
176k1384198
answered Nov 21 at 23:41
egreg
176k1384198
answered Nov 21 at 23:41
egreg
176k1384198
answered Nov 21 at 23:41
egreg
176k1384198
176k1384198
We didn't cover Taylor expansion (not even derivatives) yet, but thank you for contribution :)
– Bartosz
Nov 21 at 23:49
add a comment |
We didn't cover Taylor expansion (not even derivatives) yet, but thank you for contribution :)
– Bartosz
Nov 21 at 23:49
We didn't cover Taylor expansion (not even derivatives) yet, but thank you for contribution :)
– Bartosz
Nov 21 at 23:49
We didn't cover Taylor expansion (not even derivatives) yet, but thank you for contribution :)
– Bartosz
Nov 21 at 23:49
add a comment |
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