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Solving limit $ lim_xrightarrow 1left(fracsqrt[3]7+x^3-sqrt[2]3+x^2x-1right) $ without L'Hopital.
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I was trying all the day to resolve this problem with different method without L'Hopital but I can't do it, I would really like to up my mathematical development but the post doesn't allow me because I have less than 10 of reputation to up a image, so here is a link to one of me best try
$$ lim_xrightarrow 1left(fracsqrt[3]7+x^3-sqrt[2]3+x^2x-1right)
$$
limits-without-lhopital
edited Jul 29 at 21:46
Robert Howard
1,293620
asked Jul 29 at 21:33
mrpepo877
81
add a comment |Â
up vote
0
down vote
favorite
I was trying all the day to resolve this problem with different method without L'Hopital but I can't do it, I would really like to up my mathematical development but the post doesn't allow me because I have less than 10 of reputation to up a image, so here is a link to one of me best try
$$ lim_xrightarrow 1left(fracsqrt[3]7+x^3-sqrt[2]3+x^2x-1right)
$$
limits-without-lhopital
edited Jul 29 at 21:46
Robert Howard
1,293620
asked Jul 29 at 21:33
mrpepo877
81
Why? $qquad qquad$
– user223391
Jul 29 at 21:37
1
Regarding your try, it is false in general that $sqrta+sqrtb=sqrta+b$.
– cansomeonehelpmeout
Jul 29 at 21:37
it was a test that i did yesterday but i couldn t do that problem, so i am tring to solve now
– mrpepo877
Jul 29 at 21:38
The factorization $$(a^n-b^n)=(a-b)(a^n-1+a^n-2b+a^n-3b^2+cdots+b^n-1)$$ often helps in instances with roots. If the roots are the same, you can blindly use this. If the roots are different, you'd choose $n$ so that it is the least common multiple of the roots involved.
– Robert
Jul 29 at 21:49
2
Hint: $$lim_xto1fracroot3of7+x^3-2x-1-lim_xto1fracsqrt3+x^2-2x-1.$$ Then use $a^3-b^3=(a-b)(cdots)$ and $a^2-b^2=(a-b)(cdots)$.
– Jyrki Lahtonen
Jul 30 at 5:58
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was trying all the day to resolve this problem with different method without L'Hopital but I can't do it, I would really like to up my mathematical development but the post doesn't allow me because I have less than 10 of reputation to up a image, so here is a link to one of me best try
$$ lim_xrightarrow 1left(fracsqrt[3]7+x^3-sqrt[2]3+x^2x-1right)
$$
limits-without-lhopital
edited Jul 29 at 21:46
Robert Howard
1,293620
asked Jul 29 at 21:33
mrpepo877
81
I was trying all the day to resolve this problem with different method without L'Hopital but I can't do it, I would really like to up my mathematical development but the post doesn't allow me because I have less than 10 of reputation to up a image, so here is a link to one of me best try
$$ lim_xrightarrow 1left(fracsqrt[3]7+x^3-sqrt[2]3+x^2x-1right)
$$
limits-without-lhopital
edited Jul 29 at 21:46
Robert Howard
1,293620
asked Jul 29 at 21:33
mrpepo877
81
edited Jul 29 at 21:46
Robert Howard
1,293620
edited Jul 29 at 21:46
Robert Howard
1,293620
edited Jul 29 at 21:46
Robert Howard
1,293620
1,293620
asked Jul 29 at 21:33
mrpepo877
81
asked Jul 29 at 21:33
mrpepo877
81
asked Jul 29 at 21:33
mrpepo877
81
81
Why? $qquad qquad$
– user223391
Jul 29 at 21:37
1
Regarding your try, it is false in general that $sqrta+sqrtb=sqrta+b$.
– cansomeonehelpmeout
Jul 29 at 21:37
it was a test that i did yesterday but i couldn t do that problem, so i am tring to solve now
– mrpepo877
Jul 29 at 21:38
The factorization $$(a^n-b^n)=(a-b)(a^n-1+a^n-2b+a^n-3b^2+cdots+b^n-1)$$ often helps in instances with roots. If the roots are the same, you can blindly use this. If the roots are different, you'd choose $n$ so that it is the least common multiple of the roots involved.
– Robert
Jul 29 at 21:49
2
Hint: $$lim_xto1fracroot3of7+x^3-2x-1-lim_xto1fracsqrt3+x^2-2x-1.$$ Then use $a^3-b^3=(a-b)(cdots)$ and $a^2-b^2=(a-b)(cdots)$.
– Jyrki Lahtonen
Jul 30 at 5:58
add a comment |Â
Why? $qquad qquad$
– user223391
Jul 29 at 21:37
1
Regarding your try, it is false in general that $sqrta+sqrtb=sqrta+b$.
– cansomeonehelpmeout
Jul 29 at 21:37
it was a test that i did yesterday but i couldn t do that problem, so i am tring to solve now
– mrpepo877
Jul 29 at 21:38
The factorization $$(a^n-b^n)=(a-b)(a^n-1+a^n-2b+a^n-3b^2+cdots+b^n-1)$$ often helps in instances with roots. If the roots are the same, you can blindly use this. If the roots are different, you'd choose $n$ so that it is the least common multiple of the roots involved.
– Robert
Jul 29 at 21:49
2
Hint: $$lim_xto1fracroot3of7+x^3-2x-1-lim_xto1fracsqrt3+x^2-2x-1.$$ Then use $a^3-b^3=(a-b)(cdots)$ and $a^2-b^2=(a-b)(cdots)$.
– Jyrki Lahtonen
Jul 30 at 5:58
Why? $qquad qquad$
– user223391
Jul 29 at 21:37
Why? $qquad qquad$
– user223391
Jul 29 at 21:37
1
1
Regarding your try, it is false in general that $sqrta+sqrtb=sqrta+b$.
– cansomeonehelpmeout
Jul 29 at 21:37
Regarding your try, it is false in general that $sqrta+sqrtb=sqrta+b$.
– cansomeonehelpmeout
Jul 29 at 21:37
it was a test that i did yesterday but i couldn t do that problem, so i am tring to solve now
– mrpepo877
Jul 29 at 21:38
it was a test that i did yesterday but i couldn t do that problem, so i am tring to solve now
– mrpepo877
Jul 29 at 21:38
The factorization $$(a^n-b^n)=(a-b)(a^n-1+a^n-2b+a^n-3b^2+cdots+b^n-1)$$ often helps in instances with roots. If the roots are the same, you can blindly use this. If the roots are different, you'd choose $n$ so that it is the least common multiple of the roots involved.
– Robert
Jul 29 at 21:49
The factorization $$(a^n-b^n)=(a-b)(a^n-1+a^n-2b+a^n-3b^2+cdots+b^n-1)$$ often helps in instances with roots. If the roots are the same, you can blindly use this. If the roots are different, you'd choose $n$ so that it is the least common multiple of the roots involved.
– Robert
Jul 29 at 21:49
2
2
Hint: $$lim_xto1fracroot3of7+x^3-2x-1-lim_xto1fracsqrt3+x^2-2x-1.$$ Then use $a^3-b^3=(a-b)(cdots)$ and $a^2-b^2=(a-b)(cdots)$.
– Jyrki Lahtonen
Jul 30 at 5:58
Hint: $$lim_xto1fracroot3of7+x^3-2x-1-lim_xto1fracsqrt3+x^2-2x-1.$$ Then use $a^3-b^3=(a-b)(cdots)$ and $a^2-b^2=(a-b)(cdots)$.
– Jyrki Lahtonen
Jul 30 at 5:58
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Hint: Use the (rather lengthy, but perfectly working) identity $$a-b=fraca^6-b^6a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5$$
answered Jul 29 at 21:40
Saucy O'Path
2,393217
with $$displaystyle a=sqrt[3]7+x^3 $$ and $$displaystyle b=sqrt[2]3+x $$?
– mrpepo877
Jul 29 at 21:55
@mrpepo877 Yes, of course.
– Sobi
Jul 29 at 22:00
add a comment |Â
up vote
0
down vote
$$lim_xto1dfrac(7+x^3)^1/3-2x-1=dfracd(7+y^3)^1/3dy_(text at y=1)=?$$
answered Jul 30 at 4:37
lab bhattacharjee
214k14152263
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Hint: Use the (rather lengthy, but perfectly working) identity $$a-b=fraca^6-b^6a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5$$
answered Jul 29 at 21:40
Saucy O'Path
2,393217
with $$displaystyle a=sqrt[3]7+x^3 $$ and $$displaystyle b=sqrt[2]3+x $$?
– mrpepo877
Jul 29 at 21:55
@mrpepo877 Yes, of course.
– Sobi
Jul 29 at 22:00
add a comment |Â
up vote
4
down vote
accepted
Hint: Use the (rather lengthy, but perfectly working) identity $$a-b=fraca^6-b^6a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5$$
answered Jul 29 at 21:40
Saucy O'Path
2,393217
with $$displaystyle a=sqrt[3]7+x^3 $$ and $$displaystyle b=sqrt[2]3+x $$?
– mrpepo877
Jul 29 at 21:55
@mrpepo877 Yes, of course.
– Sobi
Jul 29 at 22:00
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Hint: Use the (rather lengthy, but perfectly working) identity $$a-b=fraca^6-b^6a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5$$
answered Jul 29 at 21:40
Saucy O'Path
2,393217
Hint: Use the (rather lengthy, but perfectly working) identity $$a-b=fraca^6-b^6a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5$$
answered Jul 29 at 21:40
Saucy O'Path
2,393217
answered Jul 29 at 21:40
Saucy O'Path
2,393217
answered Jul 29 at 21:40
Saucy O'Path
2,393217
answered Jul 29 at 21:40
Saucy O'Path
2,393217
2,393217
with $$displaystyle a=sqrt[3]7+x^3 $$ and $$displaystyle b=sqrt[2]3+x $$?
– mrpepo877
Jul 29 at 21:55
@mrpepo877 Yes, of course.
– Sobi
Jul 29 at 22:00
add a comment |Â
with $$displaystyle a=sqrt[3]7+x^3 $$ and $$displaystyle b=sqrt[2]3+x $$?
– mrpepo877
Jul 29 at 21:55
@mrpepo877 Yes, of course.
– Sobi
Jul 29 at 22:00
with $$displaystyle a=sqrt[3]7+x^3 $$ and $$displaystyle b=sqrt[2]3+x $$?
– mrpepo877
Jul 29 at 21:55
with $$displaystyle a=sqrt[3]7+x^3 $$ and $$displaystyle b=sqrt[2]3+x $$?
– mrpepo877
Jul 29 at 21:55
@mrpepo877 Yes, of course.
– Sobi
Jul 29 at 22:00
@mrpepo877 Yes, of course.
– Sobi
Jul 29 at 22:00
add a comment |Â
up vote
0
down vote
$$lim_xto1dfrac(7+x^3)^1/3-2x-1=dfracd(7+y^3)^1/3dy_(text at y=1)=?$$
answered Jul 30 at 4:37
lab bhattacharjee
214k14152263
add a comment |Â
up vote
0
down vote
$$lim_xto1dfrac(7+x^3)^1/3-2x-1=dfracd(7+y^3)^1/3dy_(text at y=1)=?$$
answered Jul 30 at 4:37
lab bhattacharjee
214k14152263
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$lim_xto1dfrac(7+x^3)^1/3-2x-1=dfracd(7+y^3)^1/3dy_(text at y=1)=?$$
answered Jul 30 at 4:37
lab bhattacharjee
214k14152263
$$lim_xto1dfrac(7+x^3)^1/3-2x-1=dfracd(7+y^3)^1/3dy_(text at y=1)=?$$
answered Jul 30 at 4:37
lab bhattacharjee
214k14152263
answered Jul 30 at 4:37
lab bhattacharjee
214k14152263
answered Jul 30 at 4:37
lab bhattacharjee
214k14152263
answered Jul 30 at 4:37
lab bhattacharjee
214k14152263
214k14152263
add a comment |Â
add a comment |Â
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Why? $qquad qquad$
– user223391
Jul 29 at 21:37
1
Regarding your try, it is false in general that $sqrta+sqrtb=sqrta+b$.
– cansomeonehelpmeout
Jul 29 at 21:37
it was a test that i did yesterday but i couldn t do that problem, so i am tring to solve now
– mrpepo877
Jul 29 at 21:38
The factorization $$(a^n-b^n)=(a-b)(a^n-1+a^n-2b+a^n-3b^2+cdots+b^n-1)$$ often helps in instances with roots. If the roots are the same, you can blindly use this. If the roots are different, you'd choose $n$ so that it is the least common multiple of the roots involved.
– Robert
Jul 29 at 21:49
2
Hint: $$lim_xto1fracroot3of7+x^3-2x-1-lim_xto1fracsqrt3+x^2-2x-1.$$ Then use $a^3-b^3=(a-b)(cdots)$ and $a^2-b^2=(a-b)(cdots)$.
– Jyrki Lahtonen
Jul 30 at 5:58