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 6 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 6 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 6 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 6 hours ago
Kevin CalderonKevin Calderon
563
edited 6 hours ago
Foobaz John
22.9k41552
asked 6 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 6 hours ago
Kevin CalderonKevin Calderon
563
asked 6 hours ago
Kevin CalderonKevin Calderon
563
asked 6 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 6 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 48 mins 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 6 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 6 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 6 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 6 hours ago
Foobaz JohnFoobaz John
22.9k41552
answered 6 hours ago
Foobaz JohnFoobaz John
22.9k41552
answered 6 hours ago
Foobaz JohnFoobaz John
22.9k41552
answered 6 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 48 mins 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 48 mins 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 48 mins 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 48 mins ago
Paras KhoslaParas Khosla
2,758423
answered 48 mins ago
Paras KhoslaParas Khosla
2,758423
answered 48 mins ago
Paras KhoslaParas Khosla
2,758423
answered 48 mins ago
Paras KhoslaParas Khosla
2,758423
2,758423
add a comment |
add a comment |
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