How to make $lim_x to 0 (1-2x)^1/x$ into $fracinftyinfty$ or $frac00$ form for L'Hospital's Rule? [duplicate] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)L'Hospital's rule for $lim_nrightarrow infty (a^n + b^n)^frac1n$.Limit Help: $lim_xtoinfty xe^-afracxln x$Finding $lim_xrightarrow 0^+fracsin(xlog(x))xlog(x)$ without L'Hospital's ruleL'Hospital's rule for $lim_nrightarrow infty (a^n + b^n)^frac1n$.L'Hospital's Rule Indeterminate Form QuestionL'Hospital's Rule and indeterminate form $fracinfty-infty$$lim_xto infty int_x^2x frac 1t dt$ using L'Hospital's ruleSolve $lim_x to 0 fracsqrt1+2x - sqrt1-4xx$ without L'Hospital's Rule.How tell that $lim_x to infty x^3e^-x^2 = 0$ using L'Hospital's Rule?How to find $lim_x to 0^+ x^sqrtx$ using L'Hospital's Rule.How to solve $lim_xrightarrow infty fracx+sin x3x+cos x$ using L'Hospital's rule
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How to make $lim_x to 0 (1-2x)^1/x$ into $fracinftyinfty$ or $frac00$ form for L'Hospital's Rule? [duplicate]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)L'Hospital's rule for $lim_nrightarrow infty (a^n + b^n)^frac1n$.Limit Help: $lim_xtoinfty xe^-afracxln x$Finding $lim_xrightarrow 0^+fracsin(xlog(x))xlog(x)$ without L'Hospital's ruleL'Hospital's rule for $lim_nrightarrow infty (a^n + b^n)^frac1n$.L'Hospital's Rule Indeterminate Form QuestionL'Hospital's Rule and indeterminate form $fracinfty-infty$$lim_xto infty int_x^2x frac 1t dt$ using L'Hospital's ruleSolve $lim_x to 0 fracsqrt1+2x - sqrt1-4xx$ without L'Hospital's Rule.How tell that $lim_x to infty x^3e^-x^2 = 0$ using L'Hospital's Rule?How to find $lim_x to 0^+ x^sqrtx$ using L'Hospital's Rule.How to solve $lim_xrightarrow infty fracx+sin x3x+cos x$ using L'Hospital's rule
$begingroup$
This question already has an answer here:
L'Hospital's rule for $lim_nrightarrow infty (a^n + b^n)^frac1n$.
3 answers
I am trying to figure out how to make $lim_x to 0 (1-2x)^1/x$ into $fracinftyinfty$ or $frac00$ form for L'Hospital's Rule.
Using Desmos.com I have found that $(1-2x)^1/x neq 1-2^1/xx^1/x$ and I'm not sure why that is either. That is the only thing I can think of to change the function's form.
calculus limits
asked Apr 8 at 19:25
LuminousNutriaLuminousNutria
57612
$endgroup$
marked as duplicate by Xander Henderson, LuminousNutria, Community♦ Apr 8 at 20:55
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
L'Hospital's rule for $lim_nrightarrow infty (a^n + b^n)^frac1n$.
3 answers
I am trying to figure out how to make $lim_x to 0 (1-2x)^1/x$ into $fracinftyinfty$ or $frac00$ form for L'Hospital's Rule.
Using Desmos.com I have found that $(1-2x)^1/x neq 1-2^1/xx^1/x$ and I'm not sure why that is either. That is the only thing I can think of to change the function's form.
calculus limits
asked Apr 8 at 19:25
LuminousNutriaLuminousNutria
57612
$endgroup$
marked as duplicate by Xander Henderson, LuminousNutria, Community♦ Apr 8 at 20:55
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Take $log$ both side
$endgroup$
– Mann
Apr 8 at 19:28
$begingroup$
math.stackexchange.com/questions/2493771/…
$endgroup$
– Xander Henderson
Apr 8 at 19:31
2
$begingroup$
It should be noted that the question @LuminousNutria marked as a dupe target is not an exact duplicate. However, I agree that it is, abstractly, a good duplicate target. The technique used there is identical: take a logarithm, work out the limit using L'H, then exponentiate.
$endgroup$
– Xander Henderson
Apr 8 at 20:48
add a comment |
$begingroup$
This question already has an answer here:
L'Hospital's rule for $lim_nrightarrow infty (a^n + b^n)^frac1n$.
3 answers
I am trying to figure out how to make $lim_x to 0 (1-2x)^1/x$ into $fracinftyinfty$ or $frac00$ form for L'Hospital's Rule.
Using Desmos.com I have found that $(1-2x)^1/x neq 1-2^1/xx^1/x$ and I'm not sure why that is either. That is the only thing I can think of to change the function's form.
calculus limits
asked Apr 8 at 19:25
LuminousNutriaLuminousNutria
57612
$endgroup$
This question already has an answer here:
L'Hospital's rule for $lim_nrightarrow infty (a^n + b^n)^frac1n$.
3 answers
I am trying to figure out how to make $lim_x to 0 (1-2x)^1/x$ into $fracinftyinfty$ or $frac00$ form for L'Hospital's Rule.
Using Desmos.com I have found that $(1-2x)^1/x neq 1-2^1/xx^1/x$ and I'm not sure why that is either. That is the only thing I can think of to change the function's form.
This question already has an answer here:
L'Hospital's rule for $lim_nrightarrow infty (a^n + b^n)^frac1n$.
3 answers
calculus limits
calculus limits
asked Apr 8 at 19:25
LuminousNutriaLuminousNutria
57612
asked Apr 8 at 19:25
LuminousNutriaLuminousNutria
57612
asked Apr 8 at 19:25
LuminousNutriaLuminousNutria
57612
asked Apr 8 at 19:25
LuminousNutriaLuminousNutria
57612
asked Apr 8 at 19:25
LuminousNutriaLuminousNutria
57612
57612
marked as duplicate by Xander Henderson, LuminousNutria, Community♦ Apr 8 at 20:55
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Xander Henderson, LuminousNutria, Community♦ Apr 8 at 20:55
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Take $log$ both side
$endgroup$
– Mann
Apr 8 at 19:28
$begingroup$
math.stackexchange.com/questions/2493771/…
$endgroup$
– Xander Henderson
Apr 8 at 19:31
2
$begingroup$
It should be noted that the question @LuminousNutria marked as a dupe target is not an exact duplicate. However, I agree that it is, abstractly, a good duplicate target. The technique used there is identical: take a logarithm, work out the limit using L'H, then exponentiate.
$endgroup$
– Xander Henderson
Apr 8 at 20:48
add a comment |
$begingroup$
Take $log$ both side
$endgroup$
– Mann
Apr 8 at 19:28
$begingroup$
math.stackexchange.com/questions/2493771/…
$endgroup$
– Xander Henderson
Apr 8 at 19:31
2
$begingroup$
It should be noted that the question @LuminousNutria marked as a dupe target is not an exact duplicate. However, I agree that it is, abstractly, a good duplicate target. The technique used there is identical: take a logarithm, work out the limit using L'H, then exponentiate.
$endgroup$
– Xander Henderson
Apr 8 at 20:48
$begingroup$
Take $log$ both side
$endgroup$
– Mann
Apr 8 at 19:28
$begingroup$
Take $log$ both side
$endgroup$
– Mann
Apr 8 at 19:28
$begingroup$
math.stackexchange.com/questions/2493771/…
$endgroup$
– Xander Henderson
Apr 8 at 19:31
$begingroup$
math.stackexchange.com/questions/2493771/…
$endgroup$
– Xander Henderson
Apr 8 at 19:31
2
2
$begingroup$
It should be noted that the question @LuminousNutria marked as a dupe target is not an exact duplicate. However, I agree that it is, abstractly, a good duplicate target. The technique used there is identical: take a logarithm, work out the limit using L'H, then exponentiate.
$endgroup$
– Xander Henderson
Apr 8 at 20:48
$begingroup$
It should be noted that the question @LuminousNutria marked as a dupe target is not an exact duplicate. However, I agree that it is, abstractly, a good duplicate target. The technique used there is identical: take a logarithm, work out the limit using L'H, then exponentiate.
$endgroup$
– Xander Henderson
Apr 8 at 20:48
add a comment |
1 Answer
1
active
oldest
votes
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A suggestion:
Calculate the limit of the logarithm first:
$$lim_xto0lnleft(1-2xright)^1/x=lim_xto 0fracln(1-2x)x.$$
If you find the limit $ell$ for the log, the limit of the expression is $mathrm e^ell$.
answered Apr 8 at 19:31
BernardBernard
124k741117
$endgroup$
$begingroup$
If I differentiate both sides I can get $frac-21-2x$ from this, which, when I replace $x$ with zero, becomes $-2$, but Symbolab.com says the answer is $frac1e^2$. Is there something else I have to do?
$endgroup$
– LuminousNutria
Apr 8 at 19:43
$begingroup$
$-2 $ is the limit of the log. Hence the limit of the expression is …
$endgroup$
– Bernard
Apr 8 at 19:47
$begingroup$
I wouldn't be able to figure that out from your hints Bernard. At least tell me a term I can look up.
$endgroup$
– LuminousNutria
Apr 8 at 19:49
2
$begingroup$
$(1-2x)^1/x=mathrm e^frac 1xln(1-2x)$ since the exponential fiunction is the inverse function of the natural logarithm..
$endgroup$
– Bernard
Apr 8 at 19:54
1
$begingroup$
Oh! I see! I get it now $lim_x to 0 (1-2x)^1/x = e^lim_x to 0 fracln(1-2x)x$ The exponent of $e$ becomes $-2$.
$endgroup$
– LuminousNutria
Apr 8 at 20:00
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A suggestion:
Calculate the limit of the logarithm first:
$$lim_xto0lnleft(1-2xright)^1/x=lim_xto 0fracln(1-2x)x.$$
If you find the limit $ell$ for the log, the limit of the expression is $mathrm e^ell$.
answered Apr 8 at 19:31
BernardBernard
124k741117
$endgroup$
$begingroup$
If I differentiate both sides I can get $frac-21-2x$ from this, which, when I replace $x$ with zero, becomes $-2$, but Symbolab.com says the answer is $frac1e^2$. Is there something else I have to do?
$endgroup$
– LuminousNutria
Apr 8 at 19:43
$begingroup$
$-2 $ is the limit of the log. Hence the limit of the expression is …
$endgroup$
– Bernard
Apr 8 at 19:47
$begingroup$
I wouldn't be able to figure that out from your hints Bernard. At least tell me a term I can look up.
$endgroup$
– LuminousNutria
Apr 8 at 19:49
2
$begingroup$
$(1-2x)^1/x=mathrm e^frac 1xln(1-2x)$ since the exponential fiunction is the inverse function of the natural logarithm..
$endgroup$
– Bernard
Apr 8 at 19:54
1
$begingroup$
Oh! I see! I get it now $lim_x to 0 (1-2x)^1/x = e^lim_x to 0 fracln(1-2x)x$ The exponent of $e$ becomes $-2$.
$endgroup$
– LuminousNutria
Apr 8 at 20:00
|
show 1 more comment
$begingroup$
A suggestion:
Calculate the limit of the logarithm first:
$$lim_xto0lnleft(1-2xright)^1/x=lim_xto 0fracln(1-2x)x.$$
If you find the limit $ell$ for the log, the limit of the expression is $mathrm e^ell$.
answered Apr 8 at 19:31
BernardBernard
124k741117
$endgroup$
$begingroup$
If I differentiate both sides I can get $frac-21-2x$ from this, which, when I replace $x$ with zero, becomes $-2$, but Symbolab.com says the answer is $frac1e^2$. Is there something else I have to do?
$endgroup$
– LuminousNutria
Apr 8 at 19:43
$begingroup$
$-2 $ is the limit of the log. Hence the limit of the expression is …
$endgroup$
– Bernard
Apr 8 at 19:47
$begingroup$
I wouldn't be able to figure that out from your hints Bernard. At least tell me a term I can look up.
$endgroup$
– LuminousNutria
Apr 8 at 19:49
2
$begingroup$
$(1-2x)^1/x=mathrm e^frac 1xln(1-2x)$ since the exponential fiunction is the inverse function of the natural logarithm..
$endgroup$
– Bernard
Apr 8 at 19:54
1
$begingroup$
Oh! I see! I get it now $lim_x to 0 (1-2x)^1/x = e^lim_x to 0 fracln(1-2x)x$ The exponent of $e$ becomes $-2$.
$endgroup$
– LuminousNutria
Apr 8 at 20:00
|
show 1 more comment
$begingroup$
A suggestion:
Calculate the limit of the logarithm first:
$$lim_xto0lnleft(1-2xright)^1/x=lim_xto 0fracln(1-2x)x.$$
If you find the limit $ell$ for the log, the limit of the expression is $mathrm e^ell$.
answered Apr 8 at 19:31
BernardBernard
124k741117
$endgroup$
A suggestion:
Calculate the limit of the logarithm first:
$$lim_xto0lnleft(1-2xright)^1/x=lim_xto 0fracln(1-2x)x.$$
If you find the limit $ell$ for the log, the limit of the expression is $mathrm e^ell$.
answered Apr 8 at 19:31
BernardBernard
124k741117
edited Apr 8 at 19:56
answered Apr 8 at 19:31
BernardBernard
124k741117
answered Apr 8 at 19:31
BernardBernard
124k741117
answered Apr 8 at 19:31
BernardBernard
124k741117
124k741117
$begingroup$
If I differentiate both sides I can get $frac-21-2x$ from this, which, when I replace $x$ with zero, becomes $-2$, but Symbolab.com says the answer is $frac1e^2$. Is there something else I have to do?
$endgroup$
– LuminousNutria
Apr 8 at 19:43
$begingroup$
$-2 $ is the limit of the log. Hence the limit of the expression is …
$endgroup$
– Bernard
Apr 8 at 19:47
$begingroup$
I wouldn't be able to figure that out from your hints Bernard. At least tell me a term I can look up.
$endgroup$
– LuminousNutria
Apr 8 at 19:49
2
$begingroup$
$(1-2x)^1/x=mathrm e^frac 1xln(1-2x)$ since the exponential fiunction is the inverse function of the natural logarithm..
$endgroup$
– Bernard
Apr 8 at 19:54
1
$begingroup$
Oh! I see! I get it now $lim_x to 0 (1-2x)^1/x = e^lim_x to 0 fracln(1-2x)x$ The exponent of $e$ becomes $-2$.
$endgroup$
– LuminousNutria
Apr 8 at 20:00
|
show 1 more comment
$begingroup$
If I differentiate both sides I can get $frac-21-2x$ from this, which, when I replace $x$ with zero, becomes $-2$, but Symbolab.com says the answer is $frac1e^2$. Is there something else I have to do?
$endgroup$
– LuminousNutria
Apr 8 at 19:43
$begingroup$
$-2 $ is the limit of the log. Hence the limit of the expression is …
$endgroup$
– Bernard
Apr 8 at 19:47
$begingroup$
I wouldn't be able to figure that out from your hints Bernard. At least tell me a term I can look up.
$endgroup$
– LuminousNutria
Apr 8 at 19:49
2
$begingroup$
$(1-2x)^1/x=mathrm e^frac 1xln(1-2x)$ since the exponential fiunction is the inverse function of the natural logarithm..
$endgroup$
– Bernard
Apr 8 at 19:54
1
$begingroup$
Oh! I see! I get it now $lim_x to 0 (1-2x)^1/x = e^lim_x to 0 fracln(1-2x)x$ The exponent of $e$ becomes $-2$.
$endgroup$
– LuminousNutria
Apr 8 at 20:00
$begingroup$
If I differentiate both sides I can get $frac-21-2x$ from this, which, when I replace $x$ with zero, becomes $-2$, but Symbolab.com says the answer is $frac1e^2$. Is there something else I have to do?
$endgroup$
– LuminousNutria
Apr 8 at 19:43
$begingroup$
If I differentiate both sides I can get $frac-21-2x$ from this, which, when I replace $x$ with zero, becomes $-2$, but Symbolab.com says the answer is $frac1e^2$. Is there something else I have to do?
$endgroup$
– LuminousNutria
Apr 8 at 19:43
$begingroup$
$-2 $ is the limit of the log. Hence the limit of the expression is …
$endgroup$
– Bernard
Apr 8 at 19:47
$begingroup$
$-2 $ is the limit of the log. Hence the limit of the expression is …
$endgroup$
– Bernard
Apr 8 at 19:47
$begingroup$
I wouldn't be able to figure that out from your hints Bernard. At least tell me a term I can look up.
$endgroup$
– LuminousNutria
Apr 8 at 19:49
$begingroup$
I wouldn't be able to figure that out from your hints Bernard. At least tell me a term I can look up.
$endgroup$
– LuminousNutria
Apr 8 at 19:49
2
2
$begingroup$
$(1-2x)^1/x=mathrm e^frac 1xln(1-2x)$ since the exponential fiunction is the inverse function of the natural logarithm..
$endgroup$
– Bernard
Apr 8 at 19:54
$begingroup$
$(1-2x)^1/x=mathrm e^frac 1xln(1-2x)$ since the exponential fiunction is the inverse function of the natural logarithm..
$endgroup$
– Bernard
Apr 8 at 19:54
1
1
$begingroup$
Oh! I see! I get it now $lim_x to 0 (1-2x)^1/x = e^lim_x to 0 fracln(1-2x)x$ The exponent of $e$ becomes $-2$.
$endgroup$
– LuminousNutria
Apr 8 at 20:00
$begingroup$
Oh! I see! I get it now $lim_x to 0 (1-2x)^1/x = e^lim_x to 0 fracln(1-2x)x$ The exponent of $e$ becomes $-2$.
$endgroup$
– LuminousNutria
Apr 8 at 20:00
|
show 1 more comment
$begingroup$
Take $log$ both side
$endgroup$
– Mann
Apr 8 at 19:28
$begingroup$
math.stackexchange.com/questions/2493771/…
$endgroup$
– Xander Henderson
Apr 8 at 19:31
2
$begingroup$
It should be noted that the question @LuminousNutria marked as a dupe target is not an exact duplicate. However, I agree that it is, abstractly, a good duplicate target. The technique used there is identical: take a logarithm, work out the limit using L'H, then exponentiate.
$endgroup$
– Xander Henderson
Apr 8 at 20:48