Proof that $exp(x+y) = exp(x)*exp(y)$ using limit definition of $exp(x)$
Multi tool use
$begingroup$
I want to prove: $exp(x+y) = exp(x)cdot exp(y)$ using the definition: $exp(x) = lim_{ntoinfty} (1+frac{x}{n})^n$
I am having trouble completing the proof, but here is my idea so far: $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = lim_{ntoinfty} left(1+frac{x}{n}right)^n cdot lim_{ntoinfty} left(1+frac{y}{n}right)^n =
lim_{ntoinfty} left(left(1+frac{x}{n}right)^n cdot left(1+frac{y}{n}right)^n right) $$
Now I rearrange the last expression: $$lim_{ntoinfty} left(1+frac{x+y+frac{xy}{n}}{n}right)^n $$
From here my idea is to somehow show that this limit is equal to $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = exp(x+y)$$ using the Squeeze Theorem and perhaps Bernoulli's Inequality, but I am at a loss as to how exactly to do it. I'd appreciate your help.
real-analysis limits exponential-function
$endgroup$
add a comment |
$begingroup$
I want to prove: $exp(x+y) = exp(x)cdot exp(y)$ using the definition: $exp(x) = lim_{ntoinfty} (1+frac{x}{n})^n$
I am having trouble completing the proof, but here is my idea so far: $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = lim_{ntoinfty} left(1+frac{x}{n}right)^n cdot lim_{ntoinfty} left(1+frac{y}{n}right)^n =
lim_{ntoinfty} left(left(1+frac{x}{n}right)^n cdot left(1+frac{y}{n}right)^n right) $$
Now I rearrange the last expression: $$lim_{ntoinfty} left(1+frac{x+y+frac{xy}{n}}{n}right)^n $$
From here my idea is to somehow show that this limit is equal to $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = exp(x+y)$$ using the Squeeze Theorem and perhaps Bernoulli's Inequality, but I am at a loss as to how exactly to do it. I'd appreciate your help.
real-analysis limits exponential-function
$endgroup$
9
$begingroup$
These should be the limit as $nrightarrowinfty$?
$endgroup$
– gd1035
Nov 27 '18 at 18:59
$begingroup$
Thank you, yes. Fixed.
$endgroup$
– TomRatTUM
Nov 27 '18 at 19:16
$begingroup$
See this answer math.stackexchange.com/a/3000717/72031
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:50
add a comment |
$begingroup$
I want to prove: $exp(x+y) = exp(x)cdot exp(y)$ using the definition: $exp(x) = lim_{ntoinfty} (1+frac{x}{n})^n$
I am having trouble completing the proof, but here is my idea so far: $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = lim_{ntoinfty} left(1+frac{x}{n}right)^n cdot lim_{ntoinfty} left(1+frac{y}{n}right)^n =
lim_{ntoinfty} left(left(1+frac{x}{n}right)^n cdot left(1+frac{y}{n}right)^n right) $$
Now I rearrange the last expression: $$lim_{ntoinfty} left(1+frac{x+y+frac{xy}{n}}{n}right)^n $$
From here my idea is to somehow show that this limit is equal to $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = exp(x+y)$$ using the Squeeze Theorem and perhaps Bernoulli's Inequality, but I am at a loss as to how exactly to do it. I'd appreciate your help.
real-analysis limits exponential-function
$endgroup$
I want to prove: $exp(x+y) = exp(x)cdot exp(y)$ using the definition: $exp(x) = lim_{ntoinfty} (1+frac{x}{n})^n$
I am having trouble completing the proof, but here is my idea so far: $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = lim_{ntoinfty} left(1+frac{x}{n}right)^n cdot lim_{ntoinfty} left(1+frac{y}{n}right)^n =
lim_{ntoinfty} left(left(1+frac{x}{n}right)^n cdot left(1+frac{y}{n}right)^n right) $$
Now I rearrange the last expression: $$lim_{ntoinfty} left(1+frac{x+y+frac{xy}{n}}{n}right)^n $$
From here my idea is to somehow show that this limit is equal to $$lim_{ntoinfty} left(1+frac{x+y}{n}right)^n = exp(x+y)$$ using the Squeeze Theorem and perhaps Bernoulli's Inequality, but I am at a loss as to how exactly to do it. I'd appreciate your help.
real-analysis limits exponential-function
real-analysis limits exponential-function
edited Nov 27 '18 at 19:15
TomRatTUM
asked Nov 27 '18 at 18:57
TomRatTUMTomRatTUM
234
234
9
$begingroup$
These should be the limit as $nrightarrowinfty$?
$endgroup$
– gd1035
Nov 27 '18 at 18:59
$begingroup$
Thank you, yes. Fixed.
$endgroup$
– TomRatTUM
Nov 27 '18 at 19:16
$begingroup$
See this answer math.stackexchange.com/a/3000717/72031
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:50
add a comment |
9
$begingroup$
These should be the limit as $nrightarrowinfty$?
$endgroup$
– gd1035
Nov 27 '18 at 18:59
$begingroup$
Thank you, yes. Fixed.
$endgroup$
– TomRatTUM
Nov 27 '18 at 19:16
$begingroup$
See this answer math.stackexchange.com/a/3000717/72031
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:50
9
9
$begingroup$
These should be the limit as $nrightarrowinfty$?
$endgroup$
– gd1035
Nov 27 '18 at 18:59
$begingroup$
These should be the limit as $nrightarrowinfty$?
$endgroup$
– gd1035
Nov 27 '18 at 18:59
$begingroup$
Thank you, yes. Fixed.
$endgroup$
– TomRatTUM
Nov 27 '18 at 19:16
$begingroup$
Thank you, yes. Fixed.
$endgroup$
– TomRatTUM
Nov 27 '18 at 19:16
$begingroup$
See this answer math.stackexchange.com/a/3000717/72031
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:50
$begingroup$
See this answer math.stackexchange.com/a/3000717/72031
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$
\dfrac{e^{x+y}}{e^x}=lim_{nto+infty}{Big(dfrac{1+frac {x+y} n}{1+frac x n}Big)^n}=
\lim_{nto+infty}Big(frac{x+y+n}{x+n}Big)^n=lim_{nto+infty}Big(1+frac y{x+n}Big)^n=
\lim_{nto+infty}frac{Big(1+frac y {x+n}Big)^{n+x}}{Big(1+frac y {x+n}Big)^x}=e^y
$$
Because $lim_{nto+infty}{Big(1+frac y {x+n}Big)^{n+x}}=e^y$ and $lim_{nto+infty}{Big(1+frac y {x+n}Big)^x}=1$
$endgroup$
$begingroup$
$x+n$ is not necessarily a positive integer.
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:52
add a comment |
$begingroup$
It suffices to show that
$$
lim_{ntoinfty}frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=1 tag{1}
$$
But
$$
frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n
$$
and as, for suitably large $n$, say $nge n_0$, we have that
$$
frac{1}{2}<1+frac{x+y}{n}<2,
$$
then
$$
left(1+frac{xy}{2n^2}right)^n<frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n<left(1+frac{2xy}{n^2}right)^n
$$
Now, for all $zinmathbb R$, we have
$$
left(1+frac{z}{n^2}right)^n=left(left(1+frac{z}{n^2}right)^{n^2}right)^{1/n}to 1,
$$
since $left(1+frac{z}{n^2}right)^{n^2}to e^z$.
Thus
$$
lim_{ntoinfty}left(1+frac{xy}{2n^2}right)^n=lim_{ntoinfty}left(1+frac{2xy}{n^2}right)^n=1,
$$
and hence $(1)$ holds.
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$
\dfrac{e^{x+y}}{e^x}=lim_{nto+infty}{Big(dfrac{1+frac {x+y} n}{1+frac x n}Big)^n}=
\lim_{nto+infty}Big(frac{x+y+n}{x+n}Big)^n=lim_{nto+infty}Big(1+frac y{x+n}Big)^n=
\lim_{nto+infty}frac{Big(1+frac y {x+n}Big)^{n+x}}{Big(1+frac y {x+n}Big)^x}=e^y
$$
Because $lim_{nto+infty}{Big(1+frac y {x+n}Big)^{n+x}}=e^y$ and $lim_{nto+infty}{Big(1+frac y {x+n}Big)^x}=1$
$endgroup$
$begingroup$
$x+n$ is not necessarily a positive integer.
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:52
add a comment |
$begingroup$
$$
\dfrac{e^{x+y}}{e^x}=lim_{nto+infty}{Big(dfrac{1+frac {x+y} n}{1+frac x n}Big)^n}=
\lim_{nto+infty}Big(frac{x+y+n}{x+n}Big)^n=lim_{nto+infty}Big(1+frac y{x+n}Big)^n=
\lim_{nto+infty}frac{Big(1+frac y {x+n}Big)^{n+x}}{Big(1+frac y {x+n}Big)^x}=e^y
$$
Because $lim_{nto+infty}{Big(1+frac y {x+n}Big)^{n+x}}=e^y$ and $lim_{nto+infty}{Big(1+frac y {x+n}Big)^x}=1$
$endgroup$
$begingroup$
$x+n$ is not necessarily a positive integer.
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:52
add a comment |
$begingroup$
$$
\dfrac{e^{x+y}}{e^x}=lim_{nto+infty}{Big(dfrac{1+frac {x+y} n}{1+frac x n}Big)^n}=
\lim_{nto+infty}Big(frac{x+y+n}{x+n}Big)^n=lim_{nto+infty}Big(1+frac y{x+n}Big)^n=
\lim_{nto+infty}frac{Big(1+frac y {x+n}Big)^{n+x}}{Big(1+frac y {x+n}Big)^x}=e^y
$$
Because $lim_{nto+infty}{Big(1+frac y {x+n}Big)^{n+x}}=e^y$ and $lim_{nto+infty}{Big(1+frac y {x+n}Big)^x}=1$
$endgroup$
$$
\dfrac{e^{x+y}}{e^x}=lim_{nto+infty}{Big(dfrac{1+frac {x+y} n}{1+frac x n}Big)^n}=
\lim_{nto+infty}Big(frac{x+y+n}{x+n}Big)^n=lim_{nto+infty}Big(1+frac y{x+n}Big)^n=
\lim_{nto+infty}frac{Big(1+frac y {x+n}Big)^{n+x}}{Big(1+frac y {x+n}Big)^x}=e^y
$$
Because $lim_{nto+infty}{Big(1+frac y {x+n}Big)^{n+x}}=e^y$ and $lim_{nto+infty}{Big(1+frac y {x+n}Big)^x}=1$
answered Nov 27 '18 at 19:06
Samvel SafaryanSamvel Safaryan
493111
493111
$begingroup$
$x+n$ is not necessarily a positive integer.
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:52
add a comment |
$begingroup$
$x+n$ is not necessarily a positive integer.
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:52
$begingroup$
$x+n$ is not necessarily a positive integer.
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:52
$begingroup$
$x+n$ is not necessarily a positive integer.
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:52
add a comment |
$begingroup$
It suffices to show that
$$
lim_{ntoinfty}frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=1 tag{1}
$$
But
$$
frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n
$$
and as, for suitably large $n$, say $nge n_0$, we have that
$$
frac{1}{2}<1+frac{x+y}{n}<2,
$$
then
$$
left(1+frac{xy}{2n^2}right)^n<frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n<left(1+frac{2xy}{n^2}right)^n
$$
Now, for all $zinmathbb R$, we have
$$
left(1+frac{z}{n^2}right)^n=left(left(1+frac{z}{n^2}right)^{n^2}right)^{1/n}to 1,
$$
since $left(1+frac{z}{n^2}right)^{n^2}to e^z$.
Thus
$$
lim_{ntoinfty}left(1+frac{xy}{2n^2}right)^n=lim_{ntoinfty}left(1+frac{2xy}{n^2}right)^n=1,
$$
and hence $(1)$ holds.
$endgroup$
add a comment |
$begingroup$
It suffices to show that
$$
lim_{ntoinfty}frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=1 tag{1}
$$
But
$$
frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n
$$
and as, for suitably large $n$, say $nge n_0$, we have that
$$
frac{1}{2}<1+frac{x+y}{n}<2,
$$
then
$$
left(1+frac{xy}{2n^2}right)^n<frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n<left(1+frac{2xy}{n^2}right)^n
$$
Now, for all $zinmathbb R$, we have
$$
left(1+frac{z}{n^2}right)^n=left(left(1+frac{z}{n^2}right)^{n^2}right)^{1/n}to 1,
$$
since $left(1+frac{z}{n^2}right)^{n^2}to e^z$.
Thus
$$
lim_{ntoinfty}left(1+frac{xy}{2n^2}right)^n=lim_{ntoinfty}left(1+frac{2xy}{n^2}right)^n=1,
$$
and hence $(1)$ holds.
$endgroup$
add a comment |
$begingroup$
It suffices to show that
$$
lim_{ntoinfty}frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=1 tag{1}
$$
But
$$
frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n
$$
and as, for suitably large $n$, say $nge n_0$, we have that
$$
frac{1}{2}<1+frac{x+y}{n}<2,
$$
then
$$
left(1+frac{xy}{2n^2}right)^n<frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n<left(1+frac{2xy}{n^2}right)^n
$$
Now, for all $zinmathbb R$, we have
$$
left(1+frac{z}{n^2}right)^n=left(left(1+frac{z}{n^2}right)^{n^2}right)^{1/n}to 1,
$$
since $left(1+frac{z}{n^2}right)^{n^2}to e^z$.
Thus
$$
lim_{ntoinfty}left(1+frac{xy}{2n^2}right)^n=lim_{ntoinfty}left(1+frac{2xy}{n^2}right)^n=1,
$$
and hence $(1)$ holds.
$endgroup$
It suffices to show that
$$
lim_{ntoinfty}frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=1 tag{1}
$$
But
$$
frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n
$$
and as, for suitably large $n$, say $nge n_0$, we have that
$$
frac{1}{2}<1+frac{x+y}{n}<2,
$$
then
$$
left(1+frac{xy}{2n^2}right)^n<frac{left(1+frac{x+y}{n}+frac{xy}{n^2}right)^n}{left(1+frac{x+y}{n}right)^n}=
left(1+frac{xy}{n^2(1+frac{x+y}{n})}right)^n<left(1+frac{2xy}{n^2}right)^n
$$
Now, for all $zinmathbb R$, we have
$$
left(1+frac{z}{n^2}right)^n=left(left(1+frac{z}{n^2}right)^{n^2}right)^{1/n}to 1,
$$
since $left(1+frac{z}{n^2}right)^{n^2}to e^z$.
Thus
$$
lim_{ntoinfty}left(1+frac{xy}{2n^2}right)^n=lim_{ntoinfty}left(1+frac{2xy}{n^2}right)^n=1,
$$
and hence $(1)$ holds.
answered Nov 27 '18 at 19:10
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.2k1384163
63.2k1384163
add a comment |
add a comment |
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9
$begingroup$
These should be the limit as $nrightarrowinfty$?
$endgroup$
– gd1035
Nov 27 '18 at 18:59
$begingroup$
Thank you, yes. Fixed.
$endgroup$
– TomRatTUM
Nov 27 '18 at 19:16
$begingroup$
See this answer math.stackexchange.com/a/3000717/72031
$endgroup$
– Paramanand Singh
Nov 28 '18 at 1:50