Solve $lim_xto 8fracx^1/3-2x-8$ [closed]
Clash Royale CLAN TAG#URR8PPP
up vote
-4
down vote
favorite
Please solve $$lim_xrightarrow 8 fracx^1/3-2x-8$$ without being indeterminate. I really don't know how...
calculus limits
edited Aug 13 at 12:56
Julián Aguirre
65k23894
asked Aug 13 at 11:03
Ree
1
closed as off-topic by amWhy, José Carlos Santos, A.Γ., Shailesh, Key Flex Aug 13 at 12:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, A.Γ., Shailesh, Key Flex
add a comment |Â
up vote
-4
down vote
favorite
Please solve $$lim_xrightarrow 8 fracx^1/3-2x-8$$ without being indeterminate. I really don't know how...
calculus limits
edited Aug 13 at 12:56
Julián Aguirre
65k23894
asked Aug 13 at 11:03
Ree
1
closed as off-topic by amWhy, José Carlos Santos, A.Γ., Shailesh, Key Flex Aug 13 at 12:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, A.Γ., Shailesh, Key Flex
2
Hint: Rationalize the numerator. But what do you mean by "without being indeterminate"?
– Rory Daulton
Aug 13 at 11:03
0/0 or infinity/infinity
– Ree
Aug 13 at 11:06
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 13 at 11:09
1
Hint: $sqrt[3]8=2$.
– A.Γ.
Aug 13 at 11:19
Use De L'Hôpital rule to get $1/12$ as the final answer.
– Von Neumann
Aug 13 at 13:04
add a comment |Â
up vote
-4
down vote
favorite
up vote
-4
down vote
favorite
Please solve $$lim_xrightarrow 8 fracx^1/3-2x-8$$ without being indeterminate. I really don't know how...
calculus limits
edited Aug 13 at 12:56
Julián Aguirre
65k23894
asked Aug 13 at 11:03
Ree
1
Please solve $$lim_xrightarrow 8 fracx^1/3-2x-8$$ without being indeterminate. I really don't know how...
calculus limits
edited Aug 13 at 12:56
Julián Aguirre
65k23894
asked Aug 13 at 11:03
Ree
1
edited Aug 13 at 12:56
Julián Aguirre
65k23894
edited Aug 13 at 12:56
Julián Aguirre
65k23894
edited Aug 13 at 12:56
Julián Aguirre
65k23894
65k23894
asked Aug 13 at 11:03
Ree
1
asked Aug 13 at 11:03
Ree
1
asked Aug 13 at 11:03
Ree
1
1
closed as off-topic by amWhy, José Carlos Santos, A.Γ., Shailesh, Key Flex Aug 13 at 12:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, A.Γ., Shailesh, Key Flex
closed as off-topic by amWhy, José Carlos Santos, A.Γ., Shailesh, Key Flex Aug 13 at 12:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, A.Γ., Shailesh, Key Flex
2
Hint: Rationalize the numerator. But what do you mean by "without being indeterminate"?
– Rory Daulton
Aug 13 at 11:03
0/0 or infinity/infinity
– Ree
Aug 13 at 11:06
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 13 at 11:09
1
Hint: $sqrt[3]8=2$.
– A.Γ.
Aug 13 at 11:19
Use De L'Hôpital rule to get $1/12$ as the final answer.
– Von Neumann
Aug 13 at 13:04
add a comment |Â
2
Hint: Rationalize the numerator. But what do you mean by "without being indeterminate"?
– Rory Daulton
Aug 13 at 11:03
0/0 or infinity/infinity
– Ree
Aug 13 at 11:06
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 13 at 11:09
1
Hint: $sqrt[3]8=2$.
– A.Γ.
Aug 13 at 11:19
Use De L'Hôpital rule to get $1/12$ as the final answer.
– Von Neumann
Aug 13 at 13:04
2
2
Hint: Rationalize the numerator. But what do you mean by "without being indeterminate"?
– Rory Daulton
Aug 13 at 11:03
Hint: Rationalize the numerator. But what do you mean by "without being indeterminate"?
– Rory Daulton
Aug 13 at 11:03
0/0 or infinity/infinity
– Ree
Aug 13 at 11:06
0/0 or infinity/infinity
– Ree
Aug 13 at 11:06
1
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 13 at 11:09
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 13 at 11:09
1
1
Hint: $sqrt[3]8=2$.
– A.Γ.
Aug 13 at 11:19
Hint: $sqrt[3]8=2$.
– A.Γ.
Aug 13 at 11:19
Use De L'Hôpital rule to get $1/12$ as the final answer.
– Von Neumann
Aug 13 at 13:04
Use De L'Hôpital rule to get $1/12$ as the final answer.
– Von Neumann
Aug 13 at 13:04
add a comment |Â
6 Answers
6
active
oldest
votes
up vote
3
down vote
Let $u=x^1/3$.
Then as $x$ approaches $8$, $u$ approaches $2$, and we have $x=u^3$, hence
beginalign*
&lim_xto8fracx^1/3-2x-8\[4pt]
=;&lim_uto 2fracu-2u^3-8\[4pt]
=;&lim_uto 2fracu-2u^3-2^3\[4pt]
=;&lim_uto 2fracu-2(u-2)(u^2+2u+4)\[4pt]
=;&lim_uto 2frac1u^2+2u+4\[4pt]
=;½^2+2(2)+4\[4pt]
=;&frac112\[4pt]
endalign*
answered Aug 13 at 11:16
quasi
33.8k22461
add a comment |Â
up vote
3
down vote
Let $x >0$:
$f(x):= x^1/3$ is differentiable.
$lim_ x rightarrow 8 dfrac f(x)-f(8)x-8= f'(8)= $
$(1/3)8^-2/3= (1/3)(1/4)=1/(12).$
answered Aug 13 at 11:41
Peter Szilas
8,0252617
Nice Peter, I don’t know why my one has been downvoted! Maybe something wrong?
– gimusi
Aug 13 at 18:00
add a comment |Â
up vote
2
down vote
Hint:
Choose $$sqrt[3]x-2=himplies x=(2+h)^3$$
As $xto8,hto0$
answered Aug 13 at 11:06
lab bhattacharjee
215k14152264
add a comment |Â
up vote
1
down vote
Notice that
$$x-8=(sqrt[3]x-2)(sqrt[3]x^2+2sqrt[3]x+4).$$
Then,
$$ lim_xrightarrow 8fracsqrt[3]x-2x-8=lim_xrightarrow 8frac1sqrt[3]x^2+2sqrt[3]x+4=frac112. $$
answered Aug 13 at 11:13
Lev Ban
53016
add a comment |Â
up vote
1
down vote
Hint: This is the rate of variation of the function $x^1/3$ from $x_0=8$:
$$fracDelta(x^1/3)Delta xBiggmvert_x_0=8stackreltextdef=fracx^1/3-8^1/3x-8.$$
answered Aug 13 at 11:14
Bernard
111k635103
add a comment |Â
up vote
-1
down vote
By l’Hopital
$$lim_xrightarrow 8 fracx^1/3-2x-8=lim_xrightarrow 8 fracfrac13x^-2/31=lim_xrightarrow 8 frac13sqrt[3]x^2=frac112$$
but any other method given in the other answers is better.
answered Aug 13 at 11:37
gimusi
66.7k73685
Gimusi.No worry.+)
– Peter Szilas
Aug 13 at 18:19
Your solution is better!
– gimusi
Aug 13 at 18:19
Not better, just slightly different:)).
– Peter Szilas
Aug 13 at 18:25
Your one is better because it is simpler!
– gimusi
Aug 13 at 18:27
add a comment |Â
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Let $u=x^1/3$.
Then as $x$ approaches $8$, $u$ approaches $2$, and we have $x=u^3$, hence
beginalign*
&lim_xto8fracx^1/3-2x-8\[4pt]
=;&lim_uto 2fracu-2u^3-8\[4pt]
=;&lim_uto 2fracu-2u^3-2^3\[4pt]
=;&lim_uto 2fracu-2(u-2)(u^2+2u+4)\[4pt]
=;&lim_uto 2frac1u^2+2u+4\[4pt]
=;½^2+2(2)+4\[4pt]
=;&frac112\[4pt]
endalign*
answered Aug 13 at 11:16
quasi
33.8k22461
add a comment |Â
up vote
3
down vote
Let $u=x^1/3$.
Then as $x$ approaches $8$, $u$ approaches $2$, and we have $x=u^3$, hence
beginalign*
&lim_xto8fracx^1/3-2x-8\[4pt]
=;&lim_uto 2fracu-2u^3-8\[4pt]
=;&lim_uto 2fracu-2u^3-2^3\[4pt]
=;&lim_uto 2fracu-2(u-2)(u^2+2u+4)\[4pt]
=;&lim_uto 2frac1u^2+2u+4\[4pt]
=;½^2+2(2)+4\[4pt]
=;&frac112\[4pt]
endalign*
answered Aug 13 at 11:16
quasi
33.8k22461
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $u=x^1/3$.
Then as $x$ approaches $8$, $u$ approaches $2$, and we have $x=u^3$, hence
beginalign*
&lim_xto8fracx^1/3-2x-8\[4pt]
=;&lim_uto 2fracu-2u^3-8\[4pt]
=;&lim_uto 2fracu-2u^3-2^3\[4pt]
=;&lim_uto 2fracu-2(u-2)(u^2+2u+4)\[4pt]
=;&lim_uto 2frac1u^2+2u+4\[4pt]
=;½^2+2(2)+4\[4pt]
=;&frac112\[4pt]
endalign*
answered Aug 13 at 11:16
quasi
33.8k22461
Let $u=x^1/3$.
Then as $x$ approaches $8$, $u$ approaches $2$, and we have $x=u^3$, hence
beginalign*
&lim_xto8fracx^1/3-2x-8\[4pt]
=;&lim_uto 2fracu-2u^3-8\[4pt]
=;&lim_uto 2fracu-2u^3-2^3\[4pt]
=;&lim_uto 2fracu-2(u-2)(u^2+2u+4)\[4pt]
=;&lim_uto 2frac1u^2+2u+4\[4pt]
=;½^2+2(2)+4\[4pt]
=;&frac112\[4pt]
endalign*
answered Aug 13 at 11:16
quasi
33.8k22461
answered Aug 13 at 11:16
quasi
33.8k22461
answered Aug 13 at 11:16
quasi
33.8k22461
answered Aug 13 at 11:16
quasi
33.8k22461
33.8k22461
add a comment |Â
add a comment |Â
up vote
3
down vote
Let $x >0$:
$f(x):= x^1/3$ is differentiable.
$lim_ x rightarrow 8 dfrac f(x)-f(8)x-8= f'(8)= $
$(1/3)8^-2/3= (1/3)(1/4)=1/(12).$
answered Aug 13 at 11:41
Peter Szilas
8,0252617
Nice Peter, I don’t know why my one has been downvoted! Maybe something wrong?
– gimusi
Aug 13 at 18:00
add a comment |Â
up vote
3
down vote
Let $x >0$:
$f(x):= x^1/3$ is differentiable.
$lim_ x rightarrow 8 dfrac f(x)-f(8)x-8= f'(8)= $
$(1/3)8^-2/3= (1/3)(1/4)=1/(12).$
answered Aug 13 at 11:41
Peter Szilas
8,0252617
Nice Peter, I don’t know why my one has been downvoted! Maybe something wrong?
– gimusi
Aug 13 at 18:00
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $x >0$:
$f(x):= x^1/3$ is differentiable.
$lim_ x rightarrow 8 dfrac f(x)-f(8)x-8= f'(8)= $
$(1/3)8^-2/3= (1/3)(1/4)=1/(12).$
answered Aug 13 at 11:41
Peter Szilas
8,0252617
Let $x >0$:
$f(x):= x^1/3$ is differentiable.
$lim_ x rightarrow 8 dfrac f(x)-f(8)x-8= f'(8)= $
$(1/3)8^-2/3= (1/3)(1/4)=1/(12).$
answered Aug 13 at 11:41
Peter Szilas
8,0252617
answered Aug 13 at 11:41
Peter Szilas
8,0252617
answered Aug 13 at 11:41
Peter Szilas
8,0252617
answered Aug 13 at 11:41
Peter Szilas
8,0252617
8,0252617
Nice Peter, I don’t know why my one has been downvoted! Maybe something wrong?
– gimusi
Aug 13 at 18:00
add a comment |Â
Nice Peter, I don’t know why my one has been downvoted! Maybe something wrong?
– gimusi
Aug 13 at 18:00
Nice Peter, I don’t know why my one has been downvoted! Maybe something wrong?
– gimusi
Aug 13 at 18:00
Nice Peter, I don’t know why my one has been downvoted! Maybe something wrong?
– gimusi
Aug 13 at 18:00
add a comment |Â
up vote
2
down vote
Hint:
Choose $$sqrt[3]x-2=himplies x=(2+h)^3$$
As $xto8,hto0$
answered Aug 13 at 11:06
lab bhattacharjee
215k14152264
add a comment |Â
up vote
2
down vote
Hint:
Choose $$sqrt[3]x-2=himplies x=(2+h)^3$$
As $xto8,hto0$
answered Aug 13 at 11:06
lab bhattacharjee
215k14152264
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint:
Choose $$sqrt[3]x-2=himplies x=(2+h)^3$$
As $xto8,hto0$
answered Aug 13 at 11:06
lab bhattacharjee
215k14152264
Hint:
Choose $$sqrt[3]x-2=himplies x=(2+h)^3$$
As $xto8,hto0$
answered Aug 13 at 11:06
lab bhattacharjee
215k14152264
answered Aug 13 at 11:06
lab bhattacharjee
215k14152264
answered Aug 13 at 11:06
lab bhattacharjee
215k14152264
answered Aug 13 at 11:06
lab bhattacharjee
215k14152264
215k14152264
add a comment |Â
add a comment |Â
up vote
1
down vote
Notice that
$$x-8=(sqrt[3]x-2)(sqrt[3]x^2+2sqrt[3]x+4).$$
Then,
$$ lim_xrightarrow 8fracsqrt[3]x-2x-8=lim_xrightarrow 8frac1sqrt[3]x^2+2sqrt[3]x+4=frac112. $$
answered Aug 13 at 11:13
Lev Ban
53016
add a comment |Â
up vote
1
down vote
Notice that
$$x-8=(sqrt[3]x-2)(sqrt[3]x^2+2sqrt[3]x+4).$$
Then,
$$ lim_xrightarrow 8fracsqrt[3]x-2x-8=lim_xrightarrow 8frac1sqrt[3]x^2+2sqrt[3]x+4=frac112. $$
answered Aug 13 at 11:13
Lev Ban
53016
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Notice that
$$x-8=(sqrt[3]x-2)(sqrt[3]x^2+2sqrt[3]x+4).$$
Then,
$$ lim_xrightarrow 8fracsqrt[3]x-2x-8=lim_xrightarrow 8frac1sqrt[3]x^2+2sqrt[3]x+4=frac112. $$
answered Aug 13 at 11:13
Lev Ban
53016
Notice that
$$x-8=(sqrt[3]x-2)(sqrt[3]x^2+2sqrt[3]x+4).$$
Then,
$$ lim_xrightarrow 8fracsqrt[3]x-2x-8=lim_xrightarrow 8frac1sqrt[3]x^2+2sqrt[3]x+4=frac112. $$
answered Aug 13 at 11:13
Lev Ban
53016
answered Aug 13 at 11:13
Lev Ban
53016
answered Aug 13 at 11:13
Lev Ban
53016
answered Aug 13 at 11:13
Lev Ban
53016
53016
add a comment |Â
add a comment |Â
up vote
1
down vote
Hint: This is the rate of variation of the function $x^1/3$ from $x_0=8$:
$$fracDelta(x^1/3)Delta xBiggmvert_x_0=8stackreltextdef=fracx^1/3-8^1/3x-8.$$
answered Aug 13 at 11:14
Bernard
111k635103
add a comment |Â
up vote
1
down vote
Hint: This is the rate of variation of the function $x^1/3$ from $x_0=8$:
$$fracDelta(x^1/3)Delta xBiggmvert_x_0=8stackreltextdef=fracx^1/3-8^1/3x-8.$$
answered Aug 13 at 11:14
Bernard
111k635103
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: This is the rate of variation of the function $x^1/3$ from $x_0=8$:
$$fracDelta(x^1/3)Delta xBiggmvert_x_0=8stackreltextdef=fracx^1/3-8^1/3x-8.$$
answered Aug 13 at 11:14
Bernard
111k635103
Hint: This is the rate of variation of the function $x^1/3$ from $x_0=8$:
$$fracDelta(x^1/3)Delta xBiggmvert_x_0=8stackreltextdef=fracx^1/3-8^1/3x-8.$$
answered Aug 13 at 11:14
Bernard
111k635103
answered Aug 13 at 11:14
Bernard
111k635103
answered Aug 13 at 11:14
Bernard
111k635103
answered Aug 13 at 11:14
Bernard
111k635103
111k635103
add a comment |Â
add a comment |Â
up vote
-1
down vote
By l’Hopital
$$lim_xrightarrow 8 fracx^1/3-2x-8=lim_xrightarrow 8 fracfrac13x^-2/31=lim_xrightarrow 8 frac13sqrt[3]x^2=frac112$$
but any other method given in the other answers is better.
answered Aug 13 at 11:37
gimusi
66.7k73685
Gimusi.No worry.+)
– Peter Szilas
Aug 13 at 18:19
Your solution is better!
– gimusi
Aug 13 at 18:19
Not better, just slightly different:)).
– Peter Szilas
Aug 13 at 18:25
Your one is better because it is simpler!
– gimusi
Aug 13 at 18:27
add a comment |Â
up vote
-1
down vote
By l’Hopital
$$lim_xrightarrow 8 fracx^1/3-2x-8=lim_xrightarrow 8 fracfrac13x^-2/31=lim_xrightarrow 8 frac13sqrt[3]x^2=frac112$$
but any other method given in the other answers is better.
answered Aug 13 at 11:37
gimusi
66.7k73685
Gimusi.No worry.+)
– Peter Szilas
Aug 13 at 18:19
Your solution is better!
– gimusi
Aug 13 at 18:19
Not better, just slightly different:)).
– Peter Szilas
Aug 13 at 18:25
Your one is better because it is simpler!
– gimusi
Aug 13 at 18:27
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
By l’Hopital
$$lim_xrightarrow 8 fracx^1/3-2x-8=lim_xrightarrow 8 fracfrac13x^-2/31=lim_xrightarrow 8 frac13sqrt[3]x^2=frac112$$
but any other method given in the other answers is better.
answered Aug 13 at 11:37
gimusi
66.7k73685
By l’Hopital
$$lim_xrightarrow 8 fracx^1/3-2x-8=lim_xrightarrow 8 fracfrac13x^-2/31=lim_xrightarrow 8 frac13sqrt[3]x^2=frac112$$
but any other method given in the other answers is better.
answered Aug 13 at 11:37
gimusi
66.7k73685
answered Aug 13 at 11:37
gimusi
66.7k73685
answered Aug 13 at 11:37
gimusi
66.7k73685
answered Aug 13 at 11:37
gimusi
66.7k73685
66.7k73685
Gimusi.No worry.+)
– Peter Szilas
Aug 13 at 18:19
Your solution is better!
– gimusi
Aug 13 at 18:19
Not better, just slightly different:)).
– Peter Szilas
Aug 13 at 18:25
Your one is better because it is simpler!
– gimusi
Aug 13 at 18:27
add a comment |Â
Gimusi.No worry.+)
– Peter Szilas
Aug 13 at 18:19
Your solution is better!
– gimusi
Aug 13 at 18:19
Not better, just slightly different:)).
– Peter Szilas
Aug 13 at 18:25
Your one is better because it is simpler!
– gimusi
Aug 13 at 18:27
Gimusi.No worry.+)
– Peter Szilas
Aug 13 at 18:19
Gimusi.No worry.+)
– Peter Szilas
Aug 13 at 18:19
Your solution is better!
– gimusi
Aug 13 at 18:19
Your solution is better!
– gimusi
Aug 13 at 18:19
Not better, just slightly different:)).
– Peter Szilas
Aug 13 at 18:25
Not better, just slightly different:)).
– Peter Szilas
Aug 13 at 18:25
Your one is better because it is simpler!
– gimusi
Aug 13 at 18:27
Your one is better because it is simpler!
– gimusi
Aug 13 at 18:27
add a comment |Â
2
Hint: Rationalize the numerator. But what do you mean by "without being indeterminate"?
– Rory Daulton
Aug 13 at 11:03
0/0 or infinity/infinity
– Ree
Aug 13 at 11:06
1
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 13 at 11:09
1
Hint: $sqrt[3]8=2$.
– A.Γ.
Aug 13 at 11:19
Use De L'Hôpital rule to get $1/12$ as the final answer.
– Von Neumann
Aug 13 at 13:04