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Whats Wrong with this approach?
Clash Royale CLAN TAG#URR8PPP
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I saw this question few moments back in here. The answer to this question is not 0 and is proceeded using Taylor expansion. I want to know where the error is in the below approach.
$lim_xto0 (frac2+cosxx^3sinx−frac3x^4)$
solving the above as follows:
$=lim_xto0(frac2x^3sinx+fraccosxx^3sinx−frac3x^4)$
$=lim_xto0(frac2x^4fracsinxx+fraccosxx^4fracsinxx−frac3x^4)$
$=lim_xto0(frac2x^4fracsinxx+frac1x^4fractanxx−frac3x^4)$
$=lim_xto0(frac2x^4lim_xto0fracsinxx+frac1x^4lim_xto0fractanxx−frac3x^4)$
$=lim_xto0(frac2x^4+frac1x^4−frac3x^4)$
$=lim_xto0(frac3x^4−frac3x^4)$
$=lim_xto0(0)$
$=0$
calculus limits
asked Aug 2 at 8:10
Ashwani Bhat
1
add a comment |Â
up vote
0
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favorite
I saw this question few moments back in here. The answer to this question is not 0 and is proceeded using Taylor expansion. I want to know where the error is in the below approach.
$lim_xto0 (frac2+cosxx^3sinx−frac3x^4)$
solving the above as follows:
$=lim_xto0(frac2x^3sinx+fraccosxx^3sinx−frac3x^4)$
$=lim_xto0(frac2x^4fracsinxx+fraccosxx^4fracsinxx−frac3x^4)$
$=lim_xto0(frac2x^4fracsinxx+frac1x^4fractanxx−frac3x^4)$
$=lim_xto0(frac2x^4lim_xto0fracsinxx+frac1x^4lim_xto0fractanxx−frac3x^4)$
$=lim_xto0(frac2x^4+frac1x^4−frac3x^4)$
$=lim_xto0(frac3x^4−frac3x^4)$
$=lim_xto0(0)$
$=0$
calculus limits
asked Aug 2 at 8:10
Ashwani Bhat
1
2
How do you go from $lim_xto0left(frac2x^4fracsin xx+frac1x^4fractan xx−frac3x^4right)$ to the next line?
– Lord Shark the Unknown
Aug 2 at 8:13
3
You cannot "move out" $x^4$ from limit, because the limit is about $x$ itself.
– Mauro ALLEGRANZA
Aug 2 at 8:14
Using individual limits on $x^4$ and $fracsinxx$
– Ashwani Bhat
Aug 2 at 8:15
3
You will have to have limit for $x^4$ also which makes the first teem infinite. You cannot write $lim frac sin x x=lim frac lim sin x x=lim frac 0 x=0$, right?
– Kavi Rama Murthy
Aug 2 at 8:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I saw this question few moments back in here. The answer to this question is not 0 and is proceeded using Taylor expansion. I want to know where the error is in the below approach.
$lim_xto0 (frac2+cosxx^3sinx−frac3x^4)$
solving the above as follows:
$=lim_xto0(frac2x^3sinx+fraccosxx^3sinx−frac3x^4)$
$=lim_xto0(frac2x^4fracsinxx+fraccosxx^4fracsinxx−frac3x^4)$
$=lim_xto0(frac2x^4fracsinxx+frac1x^4fractanxx−frac3x^4)$
$=lim_xto0(frac2x^4lim_xto0fracsinxx+frac1x^4lim_xto0fractanxx−frac3x^4)$
$=lim_xto0(frac2x^4+frac1x^4−frac3x^4)$
$=lim_xto0(frac3x^4−frac3x^4)$
$=lim_xto0(0)$
$=0$
calculus limits
asked Aug 2 at 8:10
Ashwani Bhat
1
I saw this question few moments back in here. The answer to this question is not 0 and is proceeded using Taylor expansion. I want to know where the error is in the below approach.
$lim_xto0 (frac2+cosxx^3sinx−frac3x^4)$
solving the above as follows:
$=lim_xto0(frac2x^3sinx+fraccosxx^3sinx−frac3x^4)$
$=lim_xto0(frac2x^4fracsinxx+fraccosxx^4fracsinxx−frac3x^4)$
$=lim_xto0(frac2x^4fracsinxx+frac1x^4fractanxx−frac3x^4)$
$=lim_xto0(frac2x^4lim_xto0fracsinxx+frac1x^4lim_xto0fractanxx−frac3x^4)$
$=lim_xto0(frac2x^4+frac1x^4−frac3x^4)$
$=lim_xto0(frac3x^4−frac3x^4)$
$=lim_xto0(0)$
$=0$
calculus limits
asked Aug 2 at 8:10
Ashwani Bhat
1
asked Aug 2 at 8:10
Ashwani Bhat
1
asked Aug 2 at 8:10
Ashwani Bhat
1
asked Aug 2 at 8:10
Ashwani Bhat
1
1
2
How do you go from $lim_xto0left(frac2x^4fracsin xx+frac1x^4fractan xx−frac3x^4right)$ to the next line?
– Lord Shark the Unknown
Aug 2 at 8:13
3
You cannot "move out" $x^4$ from limit, because the limit is about $x$ itself.
– Mauro ALLEGRANZA
Aug 2 at 8:14
Using individual limits on $x^4$ and $fracsinxx$
– Ashwani Bhat
Aug 2 at 8:15
3
You will have to have limit for $x^4$ also which makes the first teem infinite. You cannot write $lim frac sin x x=lim frac lim sin x x=lim frac 0 x=0$, right?
– Kavi Rama Murthy
Aug 2 at 8:24
add a comment |Â
2
How do you go from $lim_xto0left(frac2x^4fracsin xx+frac1x^4fractan xx−frac3x^4right)$ to the next line?
– Lord Shark the Unknown
Aug 2 at 8:13
3
You cannot "move out" $x^4$ from limit, because the limit is about $x$ itself.
– Mauro ALLEGRANZA
Aug 2 at 8:14
Using individual limits on $x^4$ and $fracsinxx$
– Ashwani Bhat
Aug 2 at 8:15
3
You will have to have limit for $x^4$ also which makes the first teem infinite. You cannot write $lim frac sin x x=lim frac lim sin x x=lim frac 0 x=0$, right?
– Kavi Rama Murthy
Aug 2 at 8:24
2
2
How do you go from $lim_xto0left(frac2x^4fracsin xx+frac1x^4fractan xx−frac3x^4right)$ to the next line?
– Lord Shark the Unknown
Aug 2 at 8:13
How do you go from $lim_xto0left(frac2x^4fracsin xx+frac1x^4fractan xx−frac3x^4right)$ to the next line?
– Lord Shark the Unknown
Aug 2 at 8:13
3
3
You cannot "move out" $x^4$ from limit, because the limit is about $x$ itself.
– Mauro ALLEGRANZA
Aug 2 at 8:14
You cannot "move out" $x^4$ from limit, because the limit is about $x$ itself.
– Mauro ALLEGRANZA
Aug 2 at 8:14
Using individual limits on $x^4$ and $fracsinxx$
– Ashwani Bhat
Aug 2 at 8:15
Using individual limits on $x^4$ and $fracsinxx$
– Ashwani Bhat
Aug 2 at 8:15
3
3
You will have to have limit for $x^4$ also which makes the first teem infinite. You cannot write $lim frac sin x x=lim frac lim sin x x=lim frac 0 x=0$, right?
– Kavi Rama Murthy
Aug 2 at 8:24
You will have to have limit for $x^4$ also which makes the first teem infinite. You cannot write $lim frac sin x x=lim frac lim sin x x=lim frac 0 x=0$, right?
– Kavi Rama Murthy
Aug 2 at 8:24
add a comment |Â
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2
How do you go from $lim_xto0left(frac2x^4fracsin xx+frac1x^4fractan xx−frac3x^4right)$ to the next line?
– Lord Shark the Unknown
Aug 2 at 8:13
3
You cannot "move out" $x^4$ from limit, because the limit is about $x$ itself.
– Mauro ALLEGRANZA
Aug 2 at 8:14
Using individual limits on $x^4$ and $fracsinxx$
– Ashwani Bhat
Aug 2 at 8:15
3
You will have to have limit for $x^4$ also which makes the first teem infinite. You cannot write $lim frac sin x x=lim frac lim sin x x=lim frac 0 x=0$, right?
– Kavi Rama Murthy
Aug 2 at 8:24