How to evaluate the limit where something is raised to a power of $x$?Intuitive explanation for why...
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How to evaluate the limit where something is raised to a power of $x$?
Intuitive explanation for why $left(1-frac1nright)^n to frac1e$About $lim left(1+frac {x}{n}right)^n$Finding the limit of $left( 1-frac{1}{n} right)^{n}$Showing that $lim_{n rightarrow infty} left( 1 + frac{r}{n} right)^{n} = e^{r} $.Limit of $left(1+frac{a}{x}right)^x$ with and without L'Hôpital's ruleLimit of power series $ left(1+frac{x}{m}right)^m$ with m tending to infinity$Delta x$ in limit problem?Help to evaluate this limit $lim_{x to infty}x^{frac{1}{x}}$How to evaluate this exponential fraction limit?Understanding how to evaluate $lim_{xtofracpi2} frac{2^{-cos x}-1}{x-fracpi2}$How to evaluate the following complex limit?How to evaluate integrals with infinite limitsHow to evaluate $lim_{xto infty} (x^2cdotint_{0}^x e^{t^3-x^3} dt)$Evaluate limit without L'Hopital's ruleEvaluate: $lim_limits{xto3}frac{sin(x+1)}{2x(x-3)}$Using power series to evaluate a limit
$begingroup$
I am attempting to evaluate the following limit:
$$lim_{xto infty} Biggl(frac{x+3}{x+8}Biggl)^x$$
I was wondering if anyone could share some strategies for evaluating limits raised to a power of $x$, as I have never encountered these before.
I have found the answer to be $frac{1}{e^5}$, but I am unsure how to arrive at this answer.
calculus limits
New contributor
$endgroup$
add a comment |
$begingroup$
I am attempting to evaluate the following limit:
$$lim_{xto infty} Biggl(frac{x+3}{x+8}Biggl)^x$$
I was wondering if anyone could share some strategies for evaluating limits raised to a power of $x$, as I have never encountered these before.
I have found the answer to be $frac{1}{e^5}$, but I am unsure how to arrive at this answer.
calculus limits
New contributor
$endgroup$
4
$begingroup$
Try taking the limit of the logarithm of the expression.
$endgroup$
– John Wayland Bales
yesterday
$begingroup$
This question has been asked and answered here before: (1), (2), (3), (4), (5) (just to name a few; found with Approach0.
$endgroup$
– Xander Henderson
17 hours ago
$begingroup$
Possible duplicate of About $lim left(1+frac {x}{n}right)^n$
$endgroup$
– Xander Henderson
17 hours ago
$begingroup$
It seems to be contradictory that you say you found the answer, but don't know how to find the answer.
$endgroup$
– Carl Mummert
16 hours ago
add a comment |
$begingroup$
I am attempting to evaluate the following limit:
$$lim_{xto infty} Biggl(frac{x+3}{x+8}Biggl)^x$$
I was wondering if anyone could share some strategies for evaluating limits raised to a power of $x$, as I have never encountered these before.
I have found the answer to be $frac{1}{e^5}$, but I am unsure how to arrive at this answer.
calculus limits
New contributor
$endgroup$
I am attempting to evaluate the following limit:
$$lim_{xto infty} Biggl(frac{x+3}{x+8}Biggl)^x$$
I was wondering if anyone could share some strategies for evaluating limits raised to a power of $x$, as I have never encountered these before.
I have found the answer to be $frac{1}{e^5}$, but I am unsure how to arrive at this answer.
calculus limits
calculus limits
New contributor
New contributor
edited 17 hours ago
Xander Henderson
14.8k103555
14.8k103555
New contributor
asked yesterday
jfkdasjfkjfkdasjfk
414
414
New contributor
New contributor
4
$begingroup$
Try taking the limit of the logarithm of the expression.
$endgroup$
– John Wayland Bales
yesterday
$begingroup$
This question has been asked and answered here before: (1), (2), (3), (4), (5) (just to name a few; found with Approach0.
$endgroup$
– Xander Henderson
17 hours ago
$begingroup$
Possible duplicate of About $lim left(1+frac {x}{n}right)^n$
$endgroup$
– Xander Henderson
17 hours ago
$begingroup$
It seems to be contradictory that you say you found the answer, but don't know how to find the answer.
$endgroup$
– Carl Mummert
16 hours ago
add a comment |
4
$begingroup$
Try taking the limit of the logarithm of the expression.
$endgroup$
– John Wayland Bales
yesterday
$begingroup$
This question has been asked and answered here before: (1), (2), (3), (4), (5) (just to name a few; found with Approach0.
$endgroup$
– Xander Henderson
17 hours ago
$begingroup$
Possible duplicate of About $lim left(1+frac {x}{n}right)^n$
$endgroup$
– Xander Henderson
17 hours ago
$begingroup$
It seems to be contradictory that you say you found the answer, but don't know how to find the answer.
$endgroup$
– Carl Mummert
16 hours ago
4
4
$begingroup$
Try taking the limit of the logarithm of the expression.
$endgroup$
– John Wayland Bales
yesterday
$begingroup$
Try taking the limit of the logarithm of the expression.
$endgroup$
– John Wayland Bales
yesterday
$begingroup$
This question has been asked and answered here before: (1), (2), (3), (4), (5) (just to name a few; found with Approach0.
$endgroup$
– Xander Henderson
17 hours ago
$begingroup$
This question has been asked and answered here before: (1), (2), (3), (4), (5) (just to name a few; found with Approach0.
$endgroup$
– Xander Henderson
17 hours ago
$begingroup$
Possible duplicate of About $lim left(1+frac {x}{n}right)^n$
$endgroup$
– Xander Henderson
17 hours ago
$begingroup$
Possible duplicate of About $lim left(1+frac {x}{n}right)^n$
$endgroup$
– Xander Henderson
17 hours ago
$begingroup$
It seems to be contradictory that you say you found the answer, but don't know how to find the answer.
$endgroup$
– Carl Mummert
16 hours ago
$begingroup$
It seems to be contradictory that you say you found the answer, but don't know how to find the answer.
$endgroup$
– Carl Mummert
16 hours ago
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Hint. Note that
$$Biggl({x+3over x+8}Biggl)^x=frac{(1+frac{3}{x})^x}{(1+frac{8}{x})^x}.$$
Moreover, for $anot=0$, after letting $t=x/a$ we have that
$$lim_{xto infty}(1+frac{a}{x})^x=lim_{tto infty}left(1+frac{1}{t}right)^{ta}=left(lim_{tto infty}left(1+frac{1}{t}right)^tright)^a=e^a.$$
where we used the limit which defines the Napier's constant $e$: $lim_{tto infty}left(1+frac{1}{t}right)^t=e$.
$endgroup$
$begingroup$
Thank you this helped a lot!
$endgroup$
– jfkdasjfk
yesterday
add a comment |
$begingroup$
$$=lim_{xto infty} (1-frac{5}{x})^x$$
$$=lim_{xto infty} e^{xln{(1-frac{5}{x})}}$$
$$=e^{lim_{xto infty} xln{(1-frac{5}{x})}}$$
Now in order to evaluate
$$=lim_{xto infty} xln{(1-frac{5}{x})}$$
$$=lim_{xto infty} frac{ln{(1-frac{5}{x})}}{(frac{1}{x})}$$
One can use L'Hôpitals rule giving
$$=lim_{xto infty} frac{(frac{frac{5}{x^2}}{1-frac{5}{x}})}{(-frac{1}{x^2})}$$
$$=lim_{xto infty} (-frac{5}{1-frac{5}{x}})$$
$$=-5$$
Hence the initial limit is
$$e^{-5}=frac{1}{e^5}$$
$endgroup$
add a comment |
$begingroup$
Hint:
Take the logarithm on both sides and use the fact that $exp x$ and $ln x$ are inverse functions i.e. $f(x)=exp ln f(x)$.
$$left(dfrac{x+3}{x+8}right)^x=exp xln left(dfrac{x+3}{x+8}right)$$
$endgroup$
add a comment |
$begingroup$
Take $u= x+8$. Then it's $(frac {u-5}u)^{u-8} = left ( 1-frac 5u right)^{u-8}$. Since we're taking the limit as we go to infinity, we can ignore the $-8$ $^{[1]}$. Then after another substitution $v=-frac u 5$, we have $( 1+frac 1v)^{-5v}=(( 1+frac 1v)^{v})^{-5}$, and $( 1+frac 1v)^{v}$ goes to $e$.
[1] Working it out explicitly, we have $left ( 1-frac 5u right)^{u-8}=left ( 1-frac 5u right)^{u} left ( 1-frac 5u right)^{-8}$. $left ( 1-frac 5u right)^{-8}$ goes to zero as $u$ goes to infinity, so we can eliminate that term.
$endgroup$
$begingroup$
In order to avoid mistakes like $$lim_{xto infty}left ( e^{x-8}-frac{e^x}{e^8}right)=lim_{xto infty}left ( e^{x}-frac{e^x}{e^8}right),$$ could you explain the precise meaning of "Since we're taking the limit as we go to infinity, we can ignore the $-8$"?
$endgroup$
– Pedro
yesterday
add a comment |
$begingroup$
This is an indeterminate form $1^{infty}$ (other infinite powers usually raise no difficulty).
Writing the limit as
$$lim_{xtoinfty}left(1+frac1{f(x)}right)^x$$ where $f$ tends to $infty$, we have
$$lim_{xtoinfty}left(1+frac1{f(x)}right)^x=lim_{xtoinfty}left(left(1+frac1{f(x)}right)^{f(x)}right)^{x/f(x)}=left(lim_{xtoinfty}left(1+frac1{f(x)}right)^{f(x)}right)^{lim_{xtoinfty}x/f(x)}
\=e^{lim_{xtoinfty}x/f(x)}.$$
In the given case, we have $f(x)=-(x+8)/5.$
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What about "other powers" with the form "$infty^0$"?
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– aschepler
yesterday
$begingroup$
@aschepler: I meant $a^infty$. I have rephrased.
$endgroup$
– Yves Daoust
21 hours ago
add a comment |
$begingroup$
A common strategy for limits $lim_{xto c}f(x)^{g(x)}$ of the form $1^infty$ or $infty^0$ is to compute first
$$
lim_{xto c}log(f(x)^{g(x)})=lim_{xto c}g(x)log f(x)
$$
If this limit is $l$, then the sought limit is $e^l$. In case $l=-infty$, the limit will be $0$; in case $l=infty$, the limit will be $infty$.
In the present case, a further step is useful: substitute $x=1/t$, so the limit of the logarithm becomes
$$
lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}
$$
after noticing that
$$
frac{frac{1}{t}+3}{frac{1}{t}+8}=frac{1+3t}{1+8t}
$$
This limit is easy, because it is the derivative at $0$ of $h(t)=log(1+3t)-log(1+8t)$; since
$$
h'(t)=frac{3}{1+3t}-frac{8}{1+8t}
$$
the limit is $3-8=-5$.
Alternatively, use that $log(1+u)=u+o(u)$, so you have
$$
lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}=
lim_{tto0^+}frac{3t-8t+o(t)}{t}=-5
$$
Thus your limit is $e^{-5}$.
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint. Note that
$$Biggl({x+3over x+8}Biggl)^x=frac{(1+frac{3}{x})^x}{(1+frac{8}{x})^x}.$$
Moreover, for $anot=0$, after letting $t=x/a$ we have that
$$lim_{xto infty}(1+frac{a}{x})^x=lim_{tto infty}left(1+frac{1}{t}right)^{ta}=left(lim_{tto infty}left(1+frac{1}{t}right)^tright)^a=e^a.$$
where we used the limit which defines the Napier's constant $e$: $lim_{tto infty}left(1+frac{1}{t}right)^t=e$.
$endgroup$
$begingroup$
Thank you this helped a lot!
$endgroup$
– jfkdasjfk
yesterday
add a comment |
$begingroup$
Hint. Note that
$$Biggl({x+3over x+8}Biggl)^x=frac{(1+frac{3}{x})^x}{(1+frac{8}{x})^x}.$$
Moreover, for $anot=0$, after letting $t=x/a$ we have that
$$lim_{xto infty}(1+frac{a}{x})^x=lim_{tto infty}left(1+frac{1}{t}right)^{ta}=left(lim_{tto infty}left(1+frac{1}{t}right)^tright)^a=e^a.$$
where we used the limit which defines the Napier's constant $e$: $lim_{tto infty}left(1+frac{1}{t}right)^t=e$.
$endgroup$
$begingroup$
Thank you this helped a lot!
$endgroup$
– jfkdasjfk
yesterday
add a comment |
$begingroup$
Hint. Note that
$$Biggl({x+3over x+8}Biggl)^x=frac{(1+frac{3}{x})^x}{(1+frac{8}{x})^x}.$$
Moreover, for $anot=0$, after letting $t=x/a$ we have that
$$lim_{xto infty}(1+frac{a}{x})^x=lim_{tto infty}left(1+frac{1}{t}right)^{ta}=left(lim_{tto infty}left(1+frac{1}{t}right)^tright)^a=e^a.$$
where we used the limit which defines the Napier's constant $e$: $lim_{tto infty}left(1+frac{1}{t}right)^t=e$.
$endgroup$
Hint. Note that
$$Biggl({x+3over x+8}Biggl)^x=frac{(1+frac{3}{x})^x}{(1+frac{8}{x})^x}.$$
Moreover, for $anot=0$, after letting $t=x/a$ we have that
$$lim_{xto infty}(1+frac{a}{x})^x=lim_{tto infty}left(1+frac{1}{t}right)^{ta}=left(lim_{tto infty}left(1+frac{1}{t}right)^tright)^a=e^a.$$
where we used the limit which defines the Napier's constant $e$: $lim_{tto infty}left(1+frac{1}{t}right)^t=e$.
answered yesterday
Robert ZRobert Z
99.6k1068140
99.6k1068140
$begingroup$
Thank you this helped a lot!
$endgroup$
– jfkdasjfk
yesterday
add a comment |
$begingroup$
Thank you this helped a lot!
$endgroup$
– jfkdasjfk
yesterday
$begingroup$
Thank you this helped a lot!
$endgroup$
– jfkdasjfk
yesterday
$begingroup$
Thank you this helped a lot!
$endgroup$
– jfkdasjfk
yesterday
add a comment |
$begingroup$
$$=lim_{xto infty} (1-frac{5}{x})^x$$
$$=lim_{xto infty} e^{xln{(1-frac{5}{x})}}$$
$$=e^{lim_{xto infty} xln{(1-frac{5}{x})}}$$
Now in order to evaluate
$$=lim_{xto infty} xln{(1-frac{5}{x})}$$
$$=lim_{xto infty} frac{ln{(1-frac{5}{x})}}{(frac{1}{x})}$$
One can use L'Hôpitals rule giving
$$=lim_{xto infty} frac{(frac{frac{5}{x^2}}{1-frac{5}{x}})}{(-frac{1}{x^2})}$$
$$=lim_{xto infty} (-frac{5}{1-frac{5}{x}})$$
$$=-5$$
Hence the initial limit is
$$e^{-5}=frac{1}{e^5}$$
$endgroup$
add a comment |
$begingroup$
$$=lim_{xto infty} (1-frac{5}{x})^x$$
$$=lim_{xto infty} e^{xln{(1-frac{5}{x})}}$$
$$=e^{lim_{xto infty} xln{(1-frac{5}{x})}}$$
Now in order to evaluate
$$=lim_{xto infty} xln{(1-frac{5}{x})}$$
$$=lim_{xto infty} frac{ln{(1-frac{5}{x})}}{(frac{1}{x})}$$
One can use L'Hôpitals rule giving
$$=lim_{xto infty} frac{(frac{frac{5}{x^2}}{1-frac{5}{x}})}{(-frac{1}{x^2})}$$
$$=lim_{xto infty} (-frac{5}{1-frac{5}{x}})$$
$$=-5$$
Hence the initial limit is
$$e^{-5}=frac{1}{e^5}$$
$endgroup$
add a comment |
$begingroup$
$$=lim_{xto infty} (1-frac{5}{x})^x$$
$$=lim_{xto infty} e^{xln{(1-frac{5}{x})}}$$
$$=e^{lim_{xto infty} xln{(1-frac{5}{x})}}$$
Now in order to evaluate
$$=lim_{xto infty} xln{(1-frac{5}{x})}$$
$$=lim_{xto infty} frac{ln{(1-frac{5}{x})}}{(frac{1}{x})}$$
One can use L'Hôpitals rule giving
$$=lim_{xto infty} frac{(frac{frac{5}{x^2}}{1-frac{5}{x}})}{(-frac{1}{x^2})}$$
$$=lim_{xto infty} (-frac{5}{1-frac{5}{x}})$$
$$=-5$$
Hence the initial limit is
$$e^{-5}=frac{1}{e^5}$$
$endgroup$
$$=lim_{xto infty} (1-frac{5}{x})^x$$
$$=lim_{xto infty} e^{xln{(1-frac{5}{x})}}$$
$$=e^{lim_{xto infty} xln{(1-frac{5}{x})}}$$
Now in order to evaluate
$$=lim_{xto infty} xln{(1-frac{5}{x})}$$
$$=lim_{xto infty} frac{ln{(1-frac{5}{x})}}{(frac{1}{x})}$$
One can use L'Hôpitals rule giving
$$=lim_{xto infty} frac{(frac{frac{5}{x^2}}{1-frac{5}{x}})}{(-frac{1}{x^2})}$$
$$=lim_{xto infty} (-frac{5}{1-frac{5}{x}})$$
$$=-5$$
Hence the initial limit is
$$e^{-5}=frac{1}{e^5}$$
answered yesterday
Peter ForemanPeter Foreman
2,9071214
2,9071214
add a comment |
add a comment |
$begingroup$
Hint:
Take the logarithm on both sides and use the fact that $exp x$ and $ln x$ are inverse functions i.e. $f(x)=exp ln f(x)$.
$$left(dfrac{x+3}{x+8}right)^x=exp xln left(dfrac{x+3}{x+8}right)$$
$endgroup$
add a comment |
$begingroup$
Hint:
Take the logarithm on both sides and use the fact that $exp x$ and $ln x$ are inverse functions i.e. $f(x)=exp ln f(x)$.
$$left(dfrac{x+3}{x+8}right)^x=exp xln left(dfrac{x+3}{x+8}right)$$
$endgroup$
add a comment |
$begingroup$
Hint:
Take the logarithm on both sides and use the fact that $exp x$ and $ln x$ are inverse functions i.e. $f(x)=exp ln f(x)$.
$$left(dfrac{x+3}{x+8}right)^x=exp xln left(dfrac{x+3}{x+8}right)$$
$endgroup$
Hint:
Take the logarithm on both sides and use the fact that $exp x$ and $ln x$ are inverse functions i.e. $f(x)=exp ln f(x)$.
$$left(dfrac{x+3}{x+8}right)^x=exp xln left(dfrac{x+3}{x+8}right)$$
answered yesterday
Paras KhoslaParas Khosla
1,648219
1,648219
add a comment |
add a comment |
$begingroup$
Take $u= x+8$. Then it's $(frac {u-5}u)^{u-8} = left ( 1-frac 5u right)^{u-8}$. Since we're taking the limit as we go to infinity, we can ignore the $-8$ $^{[1]}$. Then after another substitution $v=-frac u 5$, we have $( 1+frac 1v)^{-5v}=(( 1+frac 1v)^{v})^{-5}$, and $( 1+frac 1v)^{v}$ goes to $e$.
[1] Working it out explicitly, we have $left ( 1-frac 5u right)^{u-8}=left ( 1-frac 5u right)^{u} left ( 1-frac 5u right)^{-8}$. $left ( 1-frac 5u right)^{-8}$ goes to zero as $u$ goes to infinity, so we can eliminate that term.
$endgroup$
$begingroup$
In order to avoid mistakes like $$lim_{xto infty}left ( e^{x-8}-frac{e^x}{e^8}right)=lim_{xto infty}left ( e^{x}-frac{e^x}{e^8}right),$$ could you explain the precise meaning of "Since we're taking the limit as we go to infinity, we can ignore the $-8$"?
$endgroup$
– Pedro
yesterday
add a comment |
$begingroup$
Take $u= x+8$. Then it's $(frac {u-5}u)^{u-8} = left ( 1-frac 5u right)^{u-8}$. Since we're taking the limit as we go to infinity, we can ignore the $-8$ $^{[1]}$. Then after another substitution $v=-frac u 5$, we have $( 1+frac 1v)^{-5v}=(( 1+frac 1v)^{v})^{-5}$, and $( 1+frac 1v)^{v}$ goes to $e$.
[1] Working it out explicitly, we have $left ( 1-frac 5u right)^{u-8}=left ( 1-frac 5u right)^{u} left ( 1-frac 5u right)^{-8}$. $left ( 1-frac 5u right)^{-8}$ goes to zero as $u$ goes to infinity, so we can eliminate that term.
$endgroup$
$begingroup$
In order to avoid mistakes like $$lim_{xto infty}left ( e^{x-8}-frac{e^x}{e^8}right)=lim_{xto infty}left ( e^{x}-frac{e^x}{e^8}right),$$ could you explain the precise meaning of "Since we're taking the limit as we go to infinity, we can ignore the $-8$"?
$endgroup$
– Pedro
yesterday
add a comment |
$begingroup$
Take $u= x+8$. Then it's $(frac {u-5}u)^{u-8} = left ( 1-frac 5u right)^{u-8}$. Since we're taking the limit as we go to infinity, we can ignore the $-8$ $^{[1]}$. Then after another substitution $v=-frac u 5$, we have $( 1+frac 1v)^{-5v}=(( 1+frac 1v)^{v})^{-5}$, and $( 1+frac 1v)^{v}$ goes to $e$.
[1] Working it out explicitly, we have $left ( 1-frac 5u right)^{u-8}=left ( 1-frac 5u right)^{u} left ( 1-frac 5u right)^{-8}$. $left ( 1-frac 5u right)^{-8}$ goes to zero as $u$ goes to infinity, so we can eliminate that term.
$endgroup$
Take $u= x+8$. Then it's $(frac {u-5}u)^{u-8} = left ( 1-frac 5u right)^{u-8}$. Since we're taking the limit as we go to infinity, we can ignore the $-8$ $^{[1]}$. Then after another substitution $v=-frac u 5$, we have $( 1+frac 1v)^{-5v}=(( 1+frac 1v)^{v})^{-5}$, and $( 1+frac 1v)^{v}$ goes to $e$.
[1] Working it out explicitly, we have $left ( 1-frac 5u right)^{u-8}=left ( 1-frac 5u right)^{u} left ( 1-frac 5u right)^{-8}$. $left ( 1-frac 5u right)^{-8}$ goes to zero as $u$ goes to infinity, so we can eliminate that term.
edited 14 hours ago
answered yesterday
AcccumulationAcccumulation
7,0852619
7,0852619
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In order to avoid mistakes like $$lim_{xto infty}left ( e^{x-8}-frac{e^x}{e^8}right)=lim_{xto infty}left ( e^{x}-frac{e^x}{e^8}right),$$ could you explain the precise meaning of "Since we're taking the limit as we go to infinity, we can ignore the $-8$"?
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– Pedro
yesterday
add a comment |
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In order to avoid mistakes like $$lim_{xto infty}left ( e^{x-8}-frac{e^x}{e^8}right)=lim_{xto infty}left ( e^{x}-frac{e^x}{e^8}right),$$ could you explain the precise meaning of "Since we're taking the limit as we go to infinity, we can ignore the $-8$"?
$endgroup$
– Pedro
yesterday
$begingroup$
In order to avoid mistakes like $$lim_{xto infty}left ( e^{x-8}-frac{e^x}{e^8}right)=lim_{xto infty}left ( e^{x}-frac{e^x}{e^8}right),$$ could you explain the precise meaning of "Since we're taking the limit as we go to infinity, we can ignore the $-8$"?
$endgroup$
– Pedro
yesterday
$begingroup$
In order to avoid mistakes like $$lim_{xto infty}left ( e^{x-8}-frac{e^x}{e^8}right)=lim_{xto infty}left ( e^{x}-frac{e^x}{e^8}right),$$ could you explain the precise meaning of "Since we're taking the limit as we go to infinity, we can ignore the $-8$"?
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– Pedro
yesterday
add a comment |
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This is an indeterminate form $1^{infty}$ (other infinite powers usually raise no difficulty).
Writing the limit as
$$lim_{xtoinfty}left(1+frac1{f(x)}right)^x$$ where $f$ tends to $infty$, we have
$$lim_{xtoinfty}left(1+frac1{f(x)}right)^x=lim_{xtoinfty}left(left(1+frac1{f(x)}right)^{f(x)}right)^{x/f(x)}=left(lim_{xtoinfty}left(1+frac1{f(x)}right)^{f(x)}right)^{lim_{xtoinfty}x/f(x)}
\=e^{lim_{xtoinfty}x/f(x)}.$$
In the given case, we have $f(x)=-(x+8)/5.$
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What about "other powers" with the form "$infty^0$"?
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– aschepler
yesterday
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@aschepler: I meant $a^infty$. I have rephrased.
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– Yves Daoust
21 hours ago
add a comment |
$begingroup$
This is an indeterminate form $1^{infty}$ (other infinite powers usually raise no difficulty).
Writing the limit as
$$lim_{xtoinfty}left(1+frac1{f(x)}right)^x$$ where $f$ tends to $infty$, we have
$$lim_{xtoinfty}left(1+frac1{f(x)}right)^x=lim_{xtoinfty}left(left(1+frac1{f(x)}right)^{f(x)}right)^{x/f(x)}=left(lim_{xtoinfty}left(1+frac1{f(x)}right)^{f(x)}right)^{lim_{xtoinfty}x/f(x)}
\=e^{lim_{xtoinfty}x/f(x)}.$$
In the given case, we have $f(x)=-(x+8)/5.$
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What about "other powers" with the form "$infty^0$"?
$endgroup$
– aschepler
yesterday
$begingroup$
@aschepler: I meant $a^infty$. I have rephrased.
$endgroup$
– Yves Daoust
21 hours ago
add a comment |
$begingroup$
This is an indeterminate form $1^{infty}$ (other infinite powers usually raise no difficulty).
Writing the limit as
$$lim_{xtoinfty}left(1+frac1{f(x)}right)^x$$ where $f$ tends to $infty$, we have
$$lim_{xtoinfty}left(1+frac1{f(x)}right)^x=lim_{xtoinfty}left(left(1+frac1{f(x)}right)^{f(x)}right)^{x/f(x)}=left(lim_{xtoinfty}left(1+frac1{f(x)}right)^{f(x)}right)^{lim_{xtoinfty}x/f(x)}
\=e^{lim_{xtoinfty}x/f(x)}.$$
In the given case, we have $f(x)=-(x+8)/5.$
$endgroup$
This is an indeterminate form $1^{infty}$ (other infinite powers usually raise no difficulty).
Writing the limit as
$$lim_{xtoinfty}left(1+frac1{f(x)}right)^x$$ where $f$ tends to $infty$, we have
$$lim_{xtoinfty}left(1+frac1{f(x)}right)^x=lim_{xtoinfty}left(left(1+frac1{f(x)}right)^{f(x)}right)^{x/f(x)}=left(lim_{xtoinfty}left(1+frac1{f(x)}right)^{f(x)}right)^{lim_{xtoinfty}x/f(x)}
\=e^{lim_{xtoinfty}x/f(x)}.$$
In the given case, we have $f(x)=-(x+8)/5.$
edited 21 hours ago
answered yesterday
Yves DaoustYves Daoust
129k676227
129k676227
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What about "other powers" with the form "$infty^0$"?
$endgroup$
– aschepler
yesterday
$begingroup$
@aschepler: I meant $a^infty$. I have rephrased.
$endgroup$
– Yves Daoust
21 hours ago
add a comment |
$begingroup$
What about "other powers" with the form "$infty^0$"?
$endgroup$
– aschepler
yesterday
$begingroup$
@aschepler: I meant $a^infty$. I have rephrased.
$endgroup$
– Yves Daoust
21 hours ago
$begingroup$
What about "other powers" with the form "$infty^0$"?
$endgroup$
– aschepler
yesterday
$begingroup$
What about "other powers" with the form "$infty^0$"?
$endgroup$
– aschepler
yesterday
$begingroup$
@aschepler: I meant $a^infty$. I have rephrased.
$endgroup$
– Yves Daoust
21 hours ago
$begingroup$
@aschepler: I meant $a^infty$. I have rephrased.
$endgroup$
– Yves Daoust
21 hours ago
add a comment |
$begingroup$
A common strategy for limits $lim_{xto c}f(x)^{g(x)}$ of the form $1^infty$ or $infty^0$ is to compute first
$$
lim_{xto c}log(f(x)^{g(x)})=lim_{xto c}g(x)log f(x)
$$
If this limit is $l$, then the sought limit is $e^l$. In case $l=-infty$, the limit will be $0$; in case $l=infty$, the limit will be $infty$.
In the present case, a further step is useful: substitute $x=1/t$, so the limit of the logarithm becomes
$$
lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}
$$
after noticing that
$$
frac{frac{1}{t}+3}{frac{1}{t}+8}=frac{1+3t}{1+8t}
$$
This limit is easy, because it is the derivative at $0$ of $h(t)=log(1+3t)-log(1+8t)$; since
$$
h'(t)=frac{3}{1+3t}-frac{8}{1+8t}
$$
the limit is $3-8=-5$.
Alternatively, use that $log(1+u)=u+o(u)$, so you have
$$
lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}=
lim_{tto0^+}frac{3t-8t+o(t)}{t}=-5
$$
Thus your limit is $e^{-5}$.
$endgroup$
add a comment |
$begingroup$
A common strategy for limits $lim_{xto c}f(x)^{g(x)}$ of the form $1^infty$ or $infty^0$ is to compute first
$$
lim_{xto c}log(f(x)^{g(x)})=lim_{xto c}g(x)log f(x)
$$
If this limit is $l$, then the sought limit is $e^l$. In case $l=-infty$, the limit will be $0$; in case $l=infty$, the limit will be $infty$.
In the present case, a further step is useful: substitute $x=1/t$, so the limit of the logarithm becomes
$$
lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}
$$
after noticing that
$$
frac{frac{1}{t}+3}{frac{1}{t}+8}=frac{1+3t}{1+8t}
$$
This limit is easy, because it is the derivative at $0$ of $h(t)=log(1+3t)-log(1+8t)$; since
$$
h'(t)=frac{3}{1+3t}-frac{8}{1+8t}
$$
the limit is $3-8=-5$.
Alternatively, use that $log(1+u)=u+o(u)$, so you have
$$
lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}=
lim_{tto0^+}frac{3t-8t+o(t)}{t}=-5
$$
Thus your limit is $e^{-5}$.
$endgroup$
add a comment |
$begingroup$
A common strategy for limits $lim_{xto c}f(x)^{g(x)}$ of the form $1^infty$ or $infty^0$ is to compute first
$$
lim_{xto c}log(f(x)^{g(x)})=lim_{xto c}g(x)log f(x)
$$
If this limit is $l$, then the sought limit is $e^l$. In case $l=-infty$, the limit will be $0$; in case $l=infty$, the limit will be $infty$.
In the present case, a further step is useful: substitute $x=1/t$, so the limit of the logarithm becomes
$$
lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}
$$
after noticing that
$$
frac{frac{1}{t}+3}{frac{1}{t}+8}=frac{1+3t}{1+8t}
$$
This limit is easy, because it is the derivative at $0$ of $h(t)=log(1+3t)-log(1+8t)$; since
$$
h'(t)=frac{3}{1+3t}-frac{8}{1+8t}
$$
the limit is $3-8=-5$.
Alternatively, use that $log(1+u)=u+o(u)$, so you have
$$
lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}=
lim_{tto0^+}frac{3t-8t+o(t)}{t}=-5
$$
Thus your limit is $e^{-5}$.
$endgroup$
A common strategy for limits $lim_{xto c}f(x)^{g(x)}$ of the form $1^infty$ or $infty^0$ is to compute first
$$
lim_{xto c}log(f(x)^{g(x)})=lim_{xto c}g(x)log f(x)
$$
If this limit is $l$, then the sought limit is $e^l$. In case $l=-infty$, the limit will be $0$; in case $l=infty$, the limit will be $infty$.
In the present case, a further step is useful: substitute $x=1/t$, so the limit of the logarithm becomes
$$
lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}
$$
after noticing that
$$
frac{frac{1}{t}+3}{frac{1}{t}+8}=frac{1+3t}{1+8t}
$$
This limit is easy, because it is the derivative at $0$ of $h(t)=log(1+3t)-log(1+8t)$; since
$$
h'(t)=frac{3}{1+3t}-frac{8}{1+8t}
$$
the limit is $3-8=-5$.
Alternatively, use that $log(1+u)=u+o(u)$, so you have
$$
lim_{tto0^+}frac{log(1+3t)-log(1+8t)}{t}=
lim_{tto0^+}frac{3t-8t+o(t)}{t}=-5
$$
Thus your limit is $e^{-5}$.
answered 19 hours ago
egregegreg
183k1486205
183k1486205
add a comment |
add a comment |
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4
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Try taking the limit of the logarithm of the expression.
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– John Wayland Bales
yesterday
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This question has been asked and answered here before: (1), (2), (3), (4), (5) (just to name a few; found with Approach0.
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– Xander Henderson
17 hours ago
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Possible duplicate of About $lim left(1+frac {x}{n}right)^n$
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– Xander Henderson
17 hours ago
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It seems to be contradictory that you say you found the answer, but don't know how to find the answer.
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– Carl Mummert
16 hours ago