Mixing
Can't find the limit
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I'm trying to solve this limit. Wolfram showed, that there's no limit, but I can clearly see that the limit exists from graph. Tried L'Hopital's rule, but didn't get any further.
$$lim_x->-1 fracsqrt[3]1+2x+1sqrt2+x + x$$
I don't know which method should I use
limits
asked 1 hour ago
Narek Maloyan
477312
 |Â
show 3 more comments
up vote
1
down vote
favorite
I'm trying to solve this limit. Wolfram showed, that there's no limit, but I can clearly see that the limit exists from graph. Tried L'Hopital's rule, but didn't get any further.
$$lim_x->-1 fracsqrt[3]1+2x+1sqrt2+x + x$$
I don't know which method should I use
limits
asked 1 hour ago
Narek Maloyan
477312
what limit do you see on the graph?
– Rushabh Mehta
55 mins ago
@RushabhMehta, I made graph in google, so I can't say exactly, but it's around 0.4
– Narek Maloyan
52 mins ago
Could you link the graph in the comments? I am intrigued
– Rushabh Mehta
51 mins ago
@RushabhMehta, it should show google.com/…
– Narek Maloyan
50 mins ago
Well, Google is being really weird I guess. That's not even close to the actual graph of the function. This is closer.
– Rushabh Mehta
47 mins ago
 |Â
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to solve this limit. Wolfram showed, that there's no limit, but I can clearly see that the limit exists from graph. Tried L'Hopital's rule, but didn't get any further.
$$lim_x->-1 fracsqrt[3]1+2x+1sqrt2+x + x$$
I don't know which method should I use
limits
asked 1 hour ago
Narek Maloyan
477312
I'm trying to solve this limit. Wolfram showed, that there's no limit, but I can clearly see that the limit exists from graph. Tried L'Hopital's rule, but didn't get any further.
$$lim_x->-1 fracsqrt[3]1+2x+1sqrt2+x + x$$
I don't know which method should I use
limits
limits
asked 1 hour ago
Narek Maloyan
477312
asked 1 hour ago
Narek Maloyan
477312
asked 1 hour ago
Narek Maloyan
477312
asked 1 hour ago
Narek Maloyan
477312
asked 1 hour ago
Narek Maloyan
477312
477312
what limit do you see on the graph?
– Rushabh Mehta
55 mins ago
@RushabhMehta, I made graph in google, so I can't say exactly, but it's around 0.4
– Narek Maloyan
52 mins ago
Could you link the graph in the comments? I am intrigued
– Rushabh Mehta
51 mins ago
@RushabhMehta, it should show google.com/…
– Narek Maloyan
50 mins ago
Well, Google is being really weird I guess. That's not even close to the actual graph of the function. This is closer.
– Rushabh Mehta
47 mins ago
 |Â
show 3 more comments
what limit do you see on the graph?
– Rushabh Mehta
55 mins ago
@RushabhMehta, I made graph in google, so I can't say exactly, but it's around 0.4
– Narek Maloyan
52 mins ago
Could you link the graph in the comments? I am intrigued
– Rushabh Mehta
51 mins ago
@RushabhMehta, it should show google.com/…
– Narek Maloyan
50 mins ago
Well, Google is being really weird I guess. That's not even close to the actual graph of the function. This is closer.
– Rushabh Mehta
47 mins ago
what limit do you see on the graph?
– Rushabh Mehta
55 mins ago
what limit do you see on the graph?
– Rushabh Mehta
55 mins ago
@RushabhMehta, I made graph in google, so I can't say exactly, but it's around 0.4
– Narek Maloyan
52 mins ago
@RushabhMehta, I made graph in google, so I can't say exactly, but it's around 0.4
– Narek Maloyan
52 mins ago
Could you link the graph in the comments? I am intrigued
– Rushabh Mehta
51 mins ago
Could you link the graph in the comments? I am intrigued
– Rushabh Mehta
51 mins ago
@RushabhMehta, it should show google.com/…
– Narek Maloyan
50 mins ago
@RushabhMehta, it should show google.com/…
– Narek Maloyan
50 mins ago
Well, Google is being really weird I guess. That's not even close to the actual graph of the function. This is closer.
– Rushabh Mehta
47 mins ago
Well, Google is being really weird I guess. That's not even close to the actual graph of the function. This is closer.
– Rushabh Mehta
47 mins ago
 |Â
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Set $x=-1+h;$ ($hto 0$) and use the binomial approximation:
$sqrt[3]1+2(-1+h)+1=1-sqrt[3]1-2h=1-bigl(1-frac23 h+o(h)bigl)=frac23 h+o(h)$,
$sqrt2+(-1+h)-1+h=sqrt1+h-1+h=1+frac12h+o(h)-1+h=frac32h+o(h),$
so $;dfracsqrt[3]1+2x+1sqrt2+x + x=dfracfrac23 h+o(h)frac32h+o(h))=dfracfrac23+o(1)frac32+o(1)=dfrac49+o(1).$
answered 44 mins ago
Bernard
113k636104
add a comment |Â
up vote
4
down vote
Recall the formulas:
$$
beginalign
a^2 -b^2 &= (a - b)(a + b)\
a^3 + b^3 &= (a + b)(a^2 - ab + b^2)
endalign
$$
Using them we can get the following:
$$
beginalign
sqrt[3]1+2x + 1 &=
frac1 + 2x + 1sqrt[3]1+2x^2 - sqrt[3]1+2x + 1 =
frac2(x + 1)sqrt[3]1+2x^2 - sqrt[3]1+2x + 1 \
sqrt2+x + x &=
frac2 + x - x^2sqrt2+x - x =
-frac(x + 1)(x - 2)sqrt2+x - x
endalign
$$
Therefore, your limit is equal to
$$
lim_xto-1 fracsqrt[3]1+2x+1sqrt2+x + x =
lim_xto-1 -frac2(sqrt2+x - x)(x - 2)(sqrt[3]1+2x^2 - sqrt[3]1+2x + 1),
$$
which is pretty straightforward to compute.
answered 29 mins ago
mwt
814115
That's beautiful
– Narek Maloyan
27 mins ago
add a comment |Â
up vote
1
down vote
L'Hospital works:
$$lim_xto-1 fracsqrt[3]1+2x+1sqrt2+x + x=lim_xto-1 fracdfrac23(sqrt[3]1+2x)^2dfrac12sqrt2+x + 1=frac49.$$
answered 16 mins ago
Yves Daoust
116k667211
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Set $x=-1+h;$ ($hto 0$) and use the binomial approximation:
$sqrt[3]1+2(-1+h)+1=1-sqrt[3]1-2h=1-bigl(1-frac23 h+o(h)bigl)=frac23 h+o(h)$,
$sqrt2+(-1+h)-1+h=sqrt1+h-1+h=1+frac12h+o(h)-1+h=frac32h+o(h),$
so $;dfracsqrt[3]1+2x+1sqrt2+x + x=dfracfrac23 h+o(h)frac32h+o(h))=dfracfrac23+o(1)frac32+o(1)=dfrac49+o(1).$
answered 44 mins ago
Bernard
113k636104
add a comment |Â
up vote
3
down vote
accepted
Set $x=-1+h;$ ($hto 0$) and use the binomial approximation:
$sqrt[3]1+2(-1+h)+1=1-sqrt[3]1-2h=1-bigl(1-frac23 h+o(h)bigl)=frac23 h+o(h)$,
$sqrt2+(-1+h)-1+h=sqrt1+h-1+h=1+frac12h+o(h)-1+h=frac32h+o(h),$
so $;dfracsqrt[3]1+2x+1sqrt2+x + x=dfracfrac23 h+o(h)frac32h+o(h))=dfracfrac23+o(1)frac32+o(1)=dfrac49+o(1).$
answered 44 mins ago
Bernard
113k636104
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Set $x=-1+h;$ ($hto 0$) and use the binomial approximation:
$sqrt[3]1+2(-1+h)+1=1-sqrt[3]1-2h=1-bigl(1-frac23 h+o(h)bigl)=frac23 h+o(h)$,
$sqrt2+(-1+h)-1+h=sqrt1+h-1+h=1+frac12h+o(h)-1+h=frac32h+o(h),$
so $;dfracsqrt[3]1+2x+1sqrt2+x + x=dfracfrac23 h+o(h)frac32h+o(h))=dfracfrac23+o(1)frac32+o(1)=dfrac49+o(1).$
answered 44 mins ago
Bernard
113k636104
Set $x=-1+h;$ ($hto 0$) and use the binomial approximation:
$sqrt[3]1+2(-1+h)+1=1-sqrt[3]1-2h=1-bigl(1-frac23 h+o(h)bigl)=frac23 h+o(h)$,
$sqrt2+(-1+h)-1+h=sqrt1+h-1+h=1+frac12h+o(h)-1+h=frac32h+o(h),$
so $;dfracsqrt[3]1+2x+1sqrt2+x + x=dfracfrac23 h+o(h)frac32h+o(h))=dfracfrac23+o(1)frac32+o(1)=dfrac49+o(1).$
answered 44 mins ago
Bernard
113k636104
answered 44 mins ago
Bernard
113k636104
answered 44 mins ago
Bernard
113k636104
answered 44 mins ago
Bernard
113k636104
113k636104
add a comment |Â
add a comment |Â
up vote
4
down vote
Recall the formulas:
$$
beginalign
a^2 -b^2 &= (a - b)(a + b)\
a^3 + b^3 &= (a + b)(a^2 - ab + b^2)
endalign
$$
Using them we can get the following:
$$
beginalign
sqrt[3]1+2x + 1 &=
frac1 + 2x + 1sqrt[3]1+2x^2 - sqrt[3]1+2x + 1 =
frac2(x + 1)sqrt[3]1+2x^2 - sqrt[3]1+2x + 1 \
sqrt2+x + x &=
frac2 + x - x^2sqrt2+x - x =
-frac(x + 1)(x - 2)sqrt2+x - x
endalign
$$
Therefore, your limit is equal to
$$
lim_xto-1 fracsqrt[3]1+2x+1sqrt2+x + x =
lim_xto-1 -frac2(sqrt2+x - x)(x - 2)(sqrt[3]1+2x^2 - sqrt[3]1+2x + 1),
$$
which is pretty straightforward to compute.
answered 29 mins ago
mwt
814115
That's beautiful
– Narek Maloyan
27 mins ago
add a comment |Â
up vote
4
down vote
Recall the formulas:
$$
beginalign
a^2 -b^2 &= (a - b)(a + b)\
a^3 + b^3 &= (a + b)(a^2 - ab + b^2)
endalign
$$
Using them we can get the following:
$$
beginalign
sqrt[3]1+2x + 1 &=
frac1 + 2x + 1sqrt[3]1+2x^2 - sqrt[3]1+2x + 1 =
frac2(x + 1)sqrt[3]1+2x^2 - sqrt[3]1+2x + 1 \
sqrt2+x + x &=
frac2 + x - x^2sqrt2+x - x =
-frac(x + 1)(x - 2)sqrt2+x - x
endalign
$$
Therefore, your limit is equal to
$$
lim_xto-1 fracsqrt[3]1+2x+1sqrt2+x + x =
lim_xto-1 -frac2(sqrt2+x - x)(x - 2)(sqrt[3]1+2x^2 - sqrt[3]1+2x + 1),
$$
which is pretty straightforward to compute.
answered 29 mins ago
mwt
814115
That's beautiful
– Narek Maloyan
27 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Recall the formulas:
$$
beginalign
a^2 -b^2 &= (a - b)(a + b)\
a^3 + b^3 &= (a + b)(a^2 - ab + b^2)
endalign
$$
Using them we can get the following:
$$
beginalign
sqrt[3]1+2x + 1 &=
frac1 + 2x + 1sqrt[3]1+2x^2 - sqrt[3]1+2x + 1 =
frac2(x + 1)sqrt[3]1+2x^2 - sqrt[3]1+2x + 1 \
sqrt2+x + x &=
frac2 + x - x^2sqrt2+x - x =
-frac(x + 1)(x - 2)sqrt2+x - x
endalign
$$
Therefore, your limit is equal to
$$
lim_xto-1 fracsqrt[3]1+2x+1sqrt2+x + x =
lim_xto-1 -frac2(sqrt2+x - x)(x - 2)(sqrt[3]1+2x^2 - sqrt[3]1+2x + 1),
$$
which is pretty straightforward to compute.
answered 29 mins ago
mwt
814115
Recall the formulas:
$$
beginalign
a^2 -b^2 &= (a - b)(a + b)\
a^3 + b^3 &= (a + b)(a^2 - ab + b^2)
endalign
$$
Using them we can get the following:
$$
beginalign
sqrt[3]1+2x + 1 &=
frac1 + 2x + 1sqrt[3]1+2x^2 - sqrt[3]1+2x + 1 =
frac2(x + 1)sqrt[3]1+2x^2 - sqrt[3]1+2x + 1 \
sqrt2+x + x &=
frac2 + x - x^2sqrt2+x - x =
-frac(x + 1)(x - 2)sqrt2+x - x
endalign
$$
Therefore, your limit is equal to
$$
lim_xto-1 fracsqrt[3]1+2x+1sqrt2+x + x =
lim_xto-1 -frac2(sqrt2+x - x)(x - 2)(sqrt[3]1+2x^2 - sqrt[3]1+2x + 1),
$$
which is pretty straightforward to compute.
answered 29 mins ago
mwt
814115
answered 29 mins ago
mwt
814115
answered 29 mins ago
mwt
814115
answered 29 mins ago
mwt
814115
814115
That's beautiful
– Narek Maloyan
27 mins ago
add a comment |Â
That's beautiful
– Narek Maloyan
27 mins ago
That's beautiful
– Narek Maloyan
27 mins ago
That's beautiful
– Narek Maloyan
27 mins ago
add a comment |Â
up vote
1
down vote
L'Hospital works:
$$lim_xto-1 fracsqrt[3]1+2x+1sqrt2+x + x=lim_xto-1 fracdfrac23(sqrt[3]1+2x)^2dfrac12sqrt2+x + 1=frac49.$$
answered 16 mins ago
Yves Daoust
116k667211
add a comment |Â
up vote
1
down vote
L'Hospital works:
$$lim_xto-1 fracsqrt[3]1+2x+1sqrt2+x + x=lim_xto-1 fracdfrac23(sqrt[3]1+2x)^2dfrac12sqrt2+x + 1=frac49.$$
answered 16 mins ago
Yves Daoust
116k667211
add a comment |Â
up vote
1
down vote
up vote
1
down vote
L'Hospital works:
$$lim_xto-1 fracsqrt[3]1+2x+1sqrt2+x + x=lim_xto-1 fracdfrac23(sqrt[3]1+2x)^2dfrac12sqrt2+x + 1=frac49.$$
answered 16 mins ago
Yves Daoust
116k667211
L'Hospital works:
$$lim_xto-1 fracsqrt[3]1+2x+1sqrt2+x + x=lim_xto-1 fracdfrac23(sqrt[3]1+2x)^2dfrac12sqrt2+x + 1=frac49.$$
answered 16 mins ago
Yves Daoust
116k667211
answered 16 mins ago
Yves Daoust
116k667211
answered 16 mins ago
Yves Daoust
116k667211
answered 16 mins ago
Yves Daoust
116k667211
116k667211
add a comment |Â
add a comment |Â
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what limit do you see on the graph?
– Rushabh Mehta
55 mins ago
@RushabhMehta, I made graph in google, so I can't say exactly, but it's around 0.4
– Narek Maloyan
52 mins ago
Could you link the graph in the comments? I am intrigued
– Rushabh Mehta
51 mins ago
@RushabhMehta, it should show google.com/…
– Narek Maloyan
50 mins ago
Well, Google is being really weird I guess. That's not even close to the actual graph of the function. This is closer.
– Rushabh Mehta
47 mins ago