Mixing
Find the limit of $left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)$ as $n to infty$.
$begingroup$
Find the limit of $left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)$ as $n to infty$.
By expansion -
$$limlimits_{n to infty} left[1+(n^{2})(2/n) + (n^{2})(n^{2}-1)/2 dots ]/[1+2n+(2n)^{3}/3! dotsright]$$
I didn't get any result.
By applying limit directly, I'm getting indeterminate form.
How to find this limit?
limits
edited Jan 6 at 10:40
rtybase
10.7k21533
asked Jan 6 at 10:05
MathsaddictMathsaddict
3299
$endgroup$
add a comment |
$begingroup$
Find the limit of $left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)$ as $n to infty$.
By expansion -
$$limlimits_{n to infty} left[1+(n^{2})(2/n) + (n^{2})(n^{2}-1)/2 dots ]/[1+2n+(2n)^{3}/3! dotsright]$$
I didn't get any result.
By applying limit directly, I'm getting indeterminate form.
How to find this limit?
limits
edited Jan 6 at 10:40
rtybase
10.7k21533
asked Jan 6 at 10:05
MathsaddictMathsaddict
3299
$endgroup$
add a comment |
$begingroup$
Find the limit of $left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)$ as $n to infty$.
By expansion -
$$limlimits_{n to infty} left[1+(n^{2})(2/n) + (n^{2})(n^{2}-1)/2 dots ]/[1+2n+(2n)^{3}/3! dotsright]$$
I didn't get any result.
By applying limit directly, I'm getting indeterminate form.
How to find this limit?
limits
edited Jan 6 at 10:40
rtybase
10.7k21533
asked Jan 6 at 10:05
MathsaddictMathsaddict
3299
$endgroup$
Find the limit of $left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)$ as $n to infty$.
By expansion -
$$limlimits_{n to infty} left[1+(n^{2})(2/n) + (n^{2})(n^{2}-1)/2 dots ]/[1+2n+(2n)^{3}/3! dotsright]$$
I didn't get any result.
By applying limit directly, I'm getting indeterminate form.
How to find this limit?
limits
limits
edited Jan 6 at 10:40
rtybase
10.7k21533
asked Jan 6 at 10:05
MathsaddictMathsaddict
3299
edited Jan 6 at 10:40
rtybase
10.7k21533
asked Jan 6 at 10:05
MathsaddictMathsaddict
3299
edited Jan 6 at 10:40
rtybase
10.7k21533
edited Jan 6 at 10:40
rtybase
10.7k21533
edited Jan 6 at 10:40
rtybase
10.7k21533
10.7k21533
asked Jan 6 at 10:05
MathsaddictMathsaddict
3299
asked Jan 6 at 10:05
MathsaddictMathsaddict
3299
asked Jan 6 at 10:05
MathsaddictMathsaddict
3299
3299
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint. Note that
$$left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)=expleft(n^2logleft(1+ frac{2}{n}right)-2nright).$$
Now, by using the expansion $log(1+t)=t-frac{t^2}{2}+o(t^2)$ at $t=0$, we have that
$$logleft(1+ frac{2}{n}right)=frac{2}{n}-frac{2}{n^2}+o(1/n^2).$$
Can you take it from here?
answered Jan 6 at 10:07
Robert ZRobert Z
95.4k1064135
$endgroup$
1
$begingroup$
Okay, I got exp(-2). Thanks!
$endgroup$
– Mathsaddict
Jan 6 at 10:19
1
$begingroup$
Yes, that's it!
$endgroup$
– Robert Z
Jan 6 at 10:34
$begingroup$
maybe $O(1/n^3)$?
$endgroup$
– John Joy
Jan 6 at 11:08
1
$begingroup$
@JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices.
$endgroup$
– Robert Z
Jan 6 at 13:31
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint. Note that
$$left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)=expleft(n^2logleft(1+ frac{2}{n}right)-2nright).$$
Now, by using the expansion $log(1+t)=t-frac{t^2}{2}+o(t^2)$ at $t=0$, we have that
$$logleft(1+ frac{2}{n}right)=frac{2}{n}-frac{2}{n^2}+o(1/n^2).$$
Can you take it from here?
answered Jan 6 at 10:07
Robert ZRobert Z
95.4k1064135
$endgroup$
1
$begingroup$
Okay, I got exp(-2). Thanks!
$endgroup$
– Mathsaddict
Jan 6 at 10:19
1
$begingroup$
Yes, that's it!
$endgroup$
– Robert Z
Jan 6 at 10:34
$begingroup$
maybe $O(1/n^3)$?
$endgroup$
– John Joy
Jan 6 at 11:08
1
$begingroup$
@JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices.
$endgroup$
– Robert Z
Jan 6 at 13:31
add a comment |
$begingroup$
Hint. Note that
$$left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)=expleft(n^2logleft(1+ frac{2}{n}right)-2nright).$$
Now, by using the expansion $log(1+t)=t-frac{t^2}{2}+o(t^2)$ at $t=0$, we have that
$$logleft(1+ frac{2}{n}right)=frac{2}{n}-frac{2}{n^2}+o(1/n^2).$$
Can you take it from here?
answered Jan 6 at 10:07
Robert ZRobert Z
95.4k1064135
$endgroup$
1
$begingroup$
Okay, I got exp(-2). Thanks!
$endgroup$
– Mathsaddict
Jan 6 at 10:19
1
$begingroup$
Yes, that's it!
$endgroup$
– Robert Z
Jan 6 at 10:34
$begingroup$
maybe $O(1/n^3)$?
$endgroup$
– John Joy
Jan 6 at 11:08
1
$begingroup$
@JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices.
$endgroup$
– Robert Z
Jan 6 at 13:31
add a comment |
$begingroup$
Hint. Note that
$$left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)=expleft(n^2logleft(1+ frac{2}{n}right)-2nright).$$
Now, by using the expansion $log(1+t)=t-frac{t^2}{2}+o(t^2)$ at $t=0$, we have that
$$logleft(1+ frac{2}{n}right)=frac{2}{n}-frac{2}{n^2}+o(1/n^2).$$
Can you take it from here?
answered Jan 6 at 10:07
Robert ZRobert Z
95.4k1064135
$endgroup$
Hint. Note that
$$left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)=expleft(n^2logleft(1+ frac{2}{n}right)-2nright).$$
Now, by using the expansion $log(1+t)=t-frac{t^2}{2}+o(t^2)$ at $t=0$, we have that
$$logleft(1+ frac{2}{n}right)=frac{2}{n}-frac{2}{n^2}+o(1/n^2).$$
Can you take it from here?
answered Jan 6 at 10:07
Robert ZRobert Z
95.4k1064135
edited Jan 6 at 10:15
answered Jan 6 at 10:07
Robert ZRobert Z
95.4k1064135
answered Jan 6 at 10:07
Robert ZRobert Z
95.4k1064135
answered Jan 6 at 10:07
Robert ZRobert Z
95.4k1064135
95.4k1064135
1
$begingroup$
Okay, I got exp(-2). Thanks!
$endgroup$
– Mathsaddict
Jan 6 at 10:19
1
$begingroup$
Yes, that's it!
$endgroup$
– Robert Z
Jan 6 at 10:34
$begingroup$
maybe $O(1/n^3)$?
$endgroup$
– John Joy
Jan 6 at 11:08
1
$begingroup$
@JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices.
$endgroup$
– Robert Z
Jan 6 at 13:31
add a comment |
1
$begingroup$
Okay, I got exp(-2). Thanks!
$endgroup$
– Mathsaddict
Jan 6 at 10:19
1
$begingroup$
Yes, that's it!
$endgroup$
– Robert Z
Jan 6 at 10:34
$begingroup$
maybe $O(1/n^3)$?
$endgroup$
– John Joy
Jan 6 at 11:08
1
$begingroup$
@JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices.
$endgroup$
– Robert Z
Jan 6 at 13:31
1
1
$begingroup$
Okay, I got exp(-2). Thanks!
$endgroup$
– Mathsaddict
Jan 6 at 10:19
$begingroup$
Okay, I got exp(-2). Thanks!
$endgroup$
– Mathsaddict
Jan 6 at 10:19
1
1
$begingroup$
Yes, that's it!
$endgroup$
– Robert Z
Jan 6 at 10:34
$begingroup$
Yes, that's it!
$endgroup$
– Robert Z
Jan 6 at 10:34
$begingroup$
maybe $O(1/n^3)$?
$endgroup$
– John Joy
Jan 6 at 11:08
$begingroup$
maybe $O(1/n^3)$?
$endgroup$
– John Joy
Jan 6 at 11:08
1
1
$begingroup$
@JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices.
$endgroup$
– Robert Z
Jan 6 at 13:31
$begingroup$
@JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices.
$endgroup$
– Robert Z
Jan 6 at 13:31
add a comment |
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