Advance Calculus Limit question The Next CEO of Stack OverflowLimit finding of an indeterminate formI need compute a rational limit that involves rootsComplex Limit Without L'hopital'sLimit of $x^2e^x $as $x$ approaches negative infinity without using L'hopital's ruleSolving limit of radicals without L'Hopital $lim_xto 64 dfracsqrt x - 8sqrt[3] x - 4 $Solve a limit without L'Hopital: $ lim_xto0 fracln(cos5x)ln(cos7x)$Limit question - L'Hopital's rule doesn't seem to workHow can I solve this limit without L'Hopital rule?Find a limit of a function W/OUT l'Hopital's rule.Compute $lim_x rightarrow 4 frac(2x^2 - 7x -4)(-x^2 + 8x - 16)$
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Advance Calculus Limit question
The Next CEO of Stack OverflowLimit finding of an indeterminate formI need compute a rational limit that involves rootsComplex Limit Without L'hopital'sLimit of $x^2e^x $as $x$ approaches negative infinity without using L'hopital's ruleSolving limit of radicals without L'Hopital $lim_xto 64 dfracsqrt x - 8sqrt[3] x - 4 $Solve a limit without L'Hopital: $ lim_xto0 fracln(cos5x)ln(cos7x)$Limit question - L'Hopital's rule doesn't seem to workHow can I solve this limit without L'Hopital rule?Find a limit of a function W/OUT l'Hopital's rule.Compute $lim_x rightarrow 4 frac(2x^2 - 7x -4)(-x^2 + 8x - 16)$
$begingroup$
I'm trying to compute this limit without the use of L'Hopital's rule:
$$lim_x to 0^+ frac4^-1/x+4^1/x4^-1/x-4^1/x$$
I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?
calculus limits limits-without-lhopital
edited 6 hours ago
Foobaz John
22.9k41552
asked 7 hours ago
Kevin CalderonKevin Calderon
563
$endgroup$
add a comment |
$begingroup$
I'm trying to compute this limit without the use of L'Hopital's rule:
$$lim_x to 0^+ frac4^-1/x+4^1/x4^-1/x-4^1/x$$
I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?
calculus limits limits-without-lhopital
edited 6 hours ago
Foobaz John
22.9k41552
asked 7 hours ago
Kevin CalderonKevin Calderon
563
$endgroup$
add a comment |
$begingroup$
I'm trying to compute this limit without the use of L'Hopital's rule:
$$lim_x to 0^+ frac4^-1/x+4^1/x4^-1/x-4^1/x$$
I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?
calculus limits limits-without-lhopital
edited 6 hours ago
Foobaz John
22.9k41552
asked 7 hours ago
Kevin CalderonKevin Calderon
563
$endgroup$
I'm trying to compute this limit without the use of L'Hopital's rule:
$$lim_x to 0^+ frac4^-1/x+4^1/x4^-1/x-4^1/x$$
I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?
calculus limits limits-without-lhopital
calculus limits limits-without-lhopital
edited 6 hours ago
Foobaz John
22.9k41552
asked 7 hours ago
Kevin CalderonKevin Calderon
563
edited 6 hours ago
Foobaz John
22.9k41552
asked 7 hours ago
Kevin CalderonKevin Calderon
563
edited 6 hours ago
Foobaz John
22.9k41552
edited 6 hours ago
Foobaz John
22.9k41552
edited 6 hours ago
Foobaz John
22.9k41552
22.9k41552
asked 7 hours ago
Kevin CalderonKevin Calderon
563
asked 7 hours ago
Kevin CalderonKevin Calderon
563
asked 7 hours ago
Kevin CalderonKevin Calderon
563
563
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Write the limit as
$$
lim_xto 0+frac1+4^-2/x-1+4^-2/x
$$
and use the fact that
$$
lim_xto 0+frac-2x=-infty.
$$
to find that the limit equals $-1$.
answered 7 hours ago
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
add a comment |
$begingroup$
A substitution can be helpful, as it transforms the expression into a rational function:
- Set $y=4^frac1x$ and consider $y to +infty$
begineqnarray* frac4^-1/x+4^1/x4^-1/x-4^1/x
& stackrely=4^frac1x= & fracfrac1y+yfrac1y-y \
& = & fracfrac1y^2+1frac1y^2-1 \
& stackrely to +inftylongrightarrow & frac0+10-1 = -1
endeqnarray*
answered 2 hours ago
trancelocationtrancelocation
13.5k1827
$endgroup$
add a comment |
$begingroup$
$$lim_xto 0^+dfrac4^-1/x+4^1/x4^-1/x-4^1/x=lim_xto 0^+dfrac4^-2/x+14^-2/x-1$$
Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac0+10-1=-1$. Hence the required limit is $-1$.
answered 1 hour ago
Paras KhoslaParas Khosla
2,758423
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Write the limit as
$$
lim_xto 0+frac1+4^-2/x-1+4^-2/x
$$
and use the fact that
$$
lim_xto 0+frac-2x=-infty.
$$
to find that the limit equals $-1$.
answered 7 hours ago
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
add a comment |
$begingroup$
Write the limit as
$$
lim_xto 0+frac1+4^-2/x-1+4^-2/x
$$
and use the fact that
$$
lim_xto 0+frac-2x=-infty.
$$
to find that the limit equals $-1$.
answered 7 hours ago
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
add a comment |
$begingroup$
Write the limit as
$$
lim_xto 0+frac1+4^-2/x-1+4^-2/x
$$
and use the fact that
$$
lim_xto 0+frac-2x=-infty.
$$
to find that the limit equals $-1$.
answered 7 hours ago
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
Write the limit as
$$
lim_xto 0+frac1+4^-2/x-1+4^-2/x
$$
and use the fact that
$$
lim_xto 0+frac-2x=-infty.
$$
to find that the limit equals $-1$.
answered 7 hours ago
Foobaz JohnFoobaz John
22.9k41552
answered 7 hours ago
Foobaz JohnFoobaz John
22.9k41552
answered 7 hours ago
Foobaz JohnFoobaz John
22.9k41552
answered 7 hours ago
Foobaz JohnFoobaz John
22.9k41552
22.9k41552
add a comment |
add a comment |
$begingroup$
A substitution can be helpful, as it transforms the expression into a rational function:
- Set $y=4^frac1x$ and consider $y to +infty$
begineqnarray* frac4^-1/x+4^1/x4^-1/x-4^1/x
& stackrely=4^frac1x= & fracfrac1y+yfrac1y-y \
& = & fracfrac1y^2+1frac1y^2-1 \
& stackrely to +inftylongrightarrow & frac0+10-1 = -1
endeqnarray*
answered 2 hours ago
trancelocationtrancelocation
13.5k1827
$endgroup$
add a comment |
$begingroup$
A substitution can be helpful, as it transforms the expression into a rational function:
- Set $y=4^frac1x$ and consider $y to +infty$
begineqnarray* frac4^-1/x+4^1/x4^-1/x-4^1/x
& stackrely=4^frac1x= & fracfrac1y+yfrac1y-y \
& = & fracfrac1y^2+1frac1y^2-1 \
& stackrely to +inftylongrightarrow & frac0+10-1 = -1
endeqnarray*
answered 2 hours ago
trancelocationtrancelocation
13.5k1827
$endgroup$
add a comment |
$begingroup$
A substitution can be helpful, as it transforms the expression into a rational function:
- Set $y=4^frac1x$ and consider $y to +infty$
begineqnarray* frac4^-1/x+4^1/x4^-1/x-4^1/x
& stackrely=4^frac1x= & fracfrac1y+yfrac1y-y \
& = & fracfrac1y^2+1frac1y^2-1 \
& stackrely to +inftylongrightarrow & frac0+10-1 = -1
endeqnarray*
answered 2 hours ago
trancelocationtrancelocation
13.5k1827
$endgroup$
A substitution can be helpful, as it transforms the expression into a rational function:
- Set $y=4^frac1x$ and consider $y to +infty$
begineqnarray* frac4^-1/x+4^1/x4^-1/x-4^1/x
& stackrely=4^frac1x= & fracfrac1y+yfrac1y-y \
& = & fracfrac1y^2+1frac1y^2-1 \
& stackrely to +inftylongrightarrow & frac0+10-1 = -1
endeqnarray*
answered 2 hours ago
trancelocationtrancelocation
13.5k1827
answered 2 hours ago
trancelocationtrancelocation
13.5k1827
answered 2 hours ago
trancelocationtrancelocation
13.5k1827
answered 2 hours ago
trancelocationtrancelocation
13.5k1827
13.5k1827
add a comment |
add a comment |
$begingroup$
$$lim_xto 0^+dfrac4^-1/x+4^1/x4^-1/x-4^1/x=lim_xto 0^+dfrac4^-2/x+14^-2/x-1$$
Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac0+10-1=-1$. Hence the required limit is $-1$.
answered 1 hour ago
Paras KhoslaParas Khosla
2,758423
$endgroup$
add a comment |
$begingroup$
$$lim_xto 0^+dfrac4^-1/x+4^1/x4^-1/x-4^1/x=lim_xto 0^+dfrac4^-2/x+14^-2/x-1$$
Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac0+10-1=-1$. Hence the required limit is $-1$.
answered 1 hour ago
Paras KhoslaParas Khosla
2,758423
$endgroup$
add a comment |
$begingroup$
$$lim_xto 0^+dfrac4^-1/x+4^1/x4^-1/x-4^1/x=lim_xto 0^+dfrac4^-2/x+14^-2/x-1$$
Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac0+10-1=-1$. Hence the required limit is $-1$.
answered 1 hour ago
Paras KhoslaParas Khosla
2,758423
$endgroup$
$$lim_xto 0^+dfrac4^-1/x+4^1/x4^-1/x-4^1/x=lim_xto 0^+dfrac4^-2/x+14^-2/x-1$$
Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac0+10-1=-1$. Hence the required limit is $-1$.
answered 1 hour ago
Paras KhoslaParas Khosla
2,758423
answered 1 hour ago
Paras KhoslaParas Khosla
2,758423
answered 1 hour ago
Paras KhoslaParas Khosla
2,758423
answered 1 hour ago
Paras KhoslaParas Khosla
2,758423
2,758423
add a comment |
add a comment |
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