Mixing
Simple way to compute $lim_{x to 0}left(frac{1+x}{1+2x} right)^{1/x}$ without de L'Hopital and Taylor
$begingroup$
What is the simplest way you can think of to compute the following limit?
$$lim_{x to 0}left(frac{1+x}{1+2x} right)^{1/x}$$
Clearly, I can do it by using de L'Hopital or Taylor's theorems (and the properties of $exp$ and $log$), which give $$lim_{x to 0}left(frac{1+x}{1+2x} right)^{1/x} = 1/e.$$
But I'm looking for a simpler way to do it by exploiting the fact that
$$lim_{x to 0} (1+x)^{1/x} = e.$$
calculus limits limits-without-lhopital
edited Jan 12 at 18:53
Cameron Williams
22.4k43680
asked Jan 12 at 18:29
DalDal
1,50322470
$endgroup$
add a comment |
$begingroup$
What is the simplest way you can think of to compute the following limit?
$$lim_{x to 0}left(frac{1+x}{1+2x} right)^{1/x}$$
Clearly, I can do it by using de L'Hopital or Taylor's theorems (and the properties of $exp$ and $log$), which give $$lim_{x to 0}left(frac{1+x}{1+2x} right)^{1/x} = 1/e.$$
But I'm looking for a simpler way to do it by exploiting the fact that
$$lim_{x to 0} (1+x)^{1/x} = e.$$
calculus limits limits-without-lhopital
edited Jan 12 at 18:53
Cameron Williams
22.4k43680
asked Jan 12 at 18:29
DalDal
1,50322470
$endgroup$
$begingroup$
Also: $frac{1 + x}{1 + 2x} = 1 + frac{1}{frac{1 + 2x}{-x}}$
$endgroup$
– user626177
Jan 16 at 1:59
add a comment |
$begingroup$
What is the simplest way you can think of to compute the following limit?
$$lim_{x to 0}left(frac{1+x}{1+2x} right)^{1/x}$$
Clearly, I can do it by using de L'Hopital or Taylor's theorems (and the properties of $exp$ and $log$), which give $$lim_{x to 0}left(frac{1+x}{1+2x} right)^{1/x} = 1/e.$$
But I'm looking for a simpler way to do it by exploiting the fact that
$$lim_{x to 0} (1+x)^{1/x} = e.$$
calculus limits limits-without-lhopital
edited Jan 12 at 18:53
Cameron Williams
22.4k43680
asked Jan 12 at 18:29
DalDal
1,50322470
$endgroup$
What is the simplest way you can think of to compute the following limit?
$$lim_{x to 0}left(frac{1+x}{1+2x} right)^{1/x}$$
Clearly, I can do it by using de L'Hopital or Taylor's theorems (and the properties of $exp$ and $log$), which give $$lim_{x to 0}left(frac{1+x}{1+2x} right)^{1/x} = 1/e.$$
But I'm looking for a simpler way to do it by exploiting the fact that
$$lim_{x to 0} (1+x)^{1/x} = e.$$
calculus limits limits-without-lhopital
calculus limits limits-without-lhopital
edited Jan 12 at 18:53
Cameron Williams
22.4k43680
asked Jan 12 at 18:29
DalDal
1,50322470
edited Jan 12 at 18:53
Cameron Williams
22.4k43680
asked Jan 12 at 18:29
DalDal
1,50322470
edited Jan 12 at 18:53
Cameron Williams
22.4k43680
edited Jan 12 at 18:53
Cameron Williams
22.4k43680
edited Jan 12 at 18:53
Cameron Williams
22.4k43680
22.4k43680
asked Jan 12 at 18:29
DalDal
1,50322470
asked Jan 12 at 18:29
DalDal
1,50322470
asked Jan 12 at 18:29
DalDal
1,50322470
1,50322470
$begingroup$
Also: $frac{1 + x}{1 + 2x} = 1 + frac{1}{frac{1 + 2x}{-x}}$
$endgroup$
– user626177
Jan 16 at 1:59
add a comment |
$begingroup$
Also: $frac{1 + x}{1 + 2x} = 1 + frac{1}{frac{1 + 2x}{-x}}$
$endgroup$
– user626177
Jan 16 at 1:59
$begingroup$
Also: $frac{1 + x}{1 + 2x} = 1 + frac{1}{frac{1 + 2x}{-x}}$
$endgroup$
– user626177
Jan 16 at 1:59
$begingroup$
Also: $frac{1 + x}{1 + 2x} = 1 + frac{1}{frac{1 + 2x}{-x}}$
$endgroup$
– user626177
Jan 16 at 1:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$left(dfrac{1+x}{1+2x}right)^{1/x}=dfrac{(1+x)^{1/x}}{left((1+2x)^{1/2x}right)^2}.$$
edited Jan 12 at 18:51
amWhy
1
answered Jan 12 at 18:32
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
$begingroup$
Perfect. Thank you. There is just a small typo in the first line.
$endgroup$
– Dal
Jan 12 at 18:37
$begingroup$
It is very nice +1
$endgroup$
– Jimmy Sabater
Jan 12 at 18:37
add a comment |
$begingroup$
Define $y:=1/x$
Then
$$lim_{y rightarrow infty} dfrac{(1+1/y)^y}{(1+2/y)^y}=dfrac{lim_{y rightarrow infty}(1+1/y)^y}{lim_{y rightarrow infty}(1+2/y)^y}=e/e^2=e^{-1}$$
Where we used: $lim_{y rightarrow infty}(1+a/y)^y =e^a$, for $ainBbb R$.
edited Jan 12 at 21:38
John Doe
11.1k11238
answered Jan 12 at 19:27
Peter SzilasPeter Szilas
11.2k2822
$endgroup$
$begingroup$
Joe Doe.Thanks:)
$endgroup$
– Peter Szilas
Jan 12 at 22:00
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
$begingroup$
$$left(dfrac{1+x}{1+2x}right)^{1/x}=dfrac{(1+x)^{1/x}}{left((1+2x)^{1/2x}right)^2}.$$
edited Jan 12 at 18:51
amWhy
1
answered Jan 12 at 18:32
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
$begingroup$
Perfect. Thank you. There is just a small typo in the first line.
$endgroup$
– Dal
Jan 12 at 18:37
$begingroup$
It is very nice +1
$endgroup$
– Jimmy Sabater
Jan 12 at 18:37
add a comment |
$begingroup$
$$left(dfrac{1+x}{1+2x}right)^{1/x}=dfrac{(1+x)^{1/x}}{left((1+2x)^{1/2x}right)^2}.$$
edited Jan 12 at 18:51
amWhy
1
answered Jan 12 at 18:32
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
$begingroup$
Perfect. Thank you. There is just a small typo in the first line.
$endgroup$
– Dal
Jan 12 at 18:37
$begingroup$
It is very nice +1
$endgroup$
– Jimmy Sabater
Jan 12 at 18:37
add a comment |
$begingroup$
$$left(dfrac{1+x}{1+2x}right)^{1/x}=dfrac{(1+x)^{1/x}}{left((1+2x)^{1/2x}right)^2}.$$
edited Jan 12 at 18:51
amWhy
1
answered Jan 12 at 18:32
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
$$left(dfrac{1+x}{1+2x}right)^{1/x}=dfrac{(1+x)^{1/x}}{left((1+2x)^{1/2x}right)^2}.$$
edited Jan 12 at 18:51
amWhy
1
answered Jan 12 at 18:32
lab bhattacharjeelab bhattacharjee
225k15157275
edited Jan 12 at 18:51
amWhy
1
edited Jan 12 at 18:51
amWhy
1
edited Jan 12 at 18:51
amWhy
1
1
answered Jan 12 at 18:32
lab bhattacharjeelab bhattacharjee
225k15157275
answered Jan 12 at 18:32
lab bhattacharjeelab bhattacharjee
225k15157275
answered Jan 12 at 18:32
lab bhattacharjeelab bhattacharjee
225k15157275
225k15157275
$begingroup$
Perfect. Thank you. There is just a small typo in the first line.
$endgroup$
– Dal
Jan 12 at 18:37
$begingroup$
It is very nice +1
$endgroup$
– Jimmy Sabater
Jan 12 at 18:37
add a comment |
$begingroup$
Perfect. Thank you. There is just a small typo in the first line.
$endgroup$
– Dal
Jan 12 at 18:37
$begingroup$
It is very nice +1
$endgroup$
– Jimmy Sabater
Jan 12 at 18:37
$begingroup$
Perfect. Thank you. There is just a small typo in the first line.
$endgroup$
– Dal
Jan 12 at 18:37
$begingroup$
Perfect. Thank you. There is just a small typo in the first line.
$endgroup$
– Dal
Jan 12 at 18:37
$begingroup$
It is very nice +1
$endgroup$
– Jimmy Sabater
Jan 12 at 18:37
$begingroup$
It is very nice +1
$endgroup$
– Jimmy Sabater
Jan 12 at 18:37
add a comment |
$begingroup$
Define $y:=1/x$
Then
$$lim_{y rightarrow infty} dfrac{(1+1/y)^y}{(1+2/y)^y}=dfrac{lim_{y rightarrow infty}(1+1/y)^y}{lim_{y rightarrow infty}(1+2/y)^y}=e/e^2=e^{-1}$$
Where we used: $lim_{y rightarrow infty}(1+a/y)^y =e^a$, for $ainBbb R$.
edited Jan 12 at 21:38
John Doe
11.1k11238
answered Jan 12 at 19:27
Peter SzilasPeter Szilas
11.2k2822
$endgroup$
$begingroup$
Joe Doe.Thanks:)
$endgroup$
– Peter Szilas
Jan 12 at 22:00
add a comment |
$begingroup$
Define $y:=1/x$
Then
$$lim_{y rightarrow infty} dfrac{(1+1/y)^y}{(1+2/y)^y}=dfrac{lim_{y rightarrow infty}(1+1/y)^y}{lim_{y rightarrow infty}(1+2/y)^y}=e/e^2=e^{-1}$$
Where we used: $lim_{y rightarrow infty}(1+a/y)^y =e^a$, for $ainBbb R$.
edited Jan 12 at 21:38
John Doe
11.1k11238
answered Jan 12 at 19:27
Peter SzilasPeter Szilas
11.2k2822
$endgroup$
$begingroup$
Joe Doe.Thanks:)
$endgroup$
– Peter Szilas
Jan 12 at 22:00
add a comment |
$begingroup$
Define $y:=1/x$
Then
$$lim_{y rightarrow infty} dfrac{(1+1/y)^y}{(1+2/y)^y}=dfrac{lim_{y rightarrow infty}(1+1/y)^y}{lim_{y rightarrow infty}(1+2/y)^y}=e/e^2=e^{-1}$$
Where we used: $lim_{y rightarrow infty}(1+a/y)^y =e^a$, for $ainBbb R$.
edited Jan 12 at 21:38
John Doe
11.1k11238
answered Jan 12 at 19:27
Peter SzilasPeter Szilas
11.2k2822
$endgroup$
Define $y:=1/x$
Then
$$lim_{y rightarrow infty} dfrac{(1+1/y)^y}{(1+2/y)^y}=dfrac{lim_{y rightarrow infty}(1+1/y)^y}{lim_{y rightarrow infty}(1+2/y)^y}=e/e^2=e^{-1}$$
Where we used: $lim_{y rightarrow infty}(1+a/y)^y =e^a$, for $ainBbb R$.
edited Jan 12 at 21:38
John Doe
11.1k11238
answered Jan 12 at 19:27
Peter SzilasPeter Szilas
11.2k2822
edited Jan 12 at 21:38
John Doe
11.1k11238
edited Jan 12 at 21:38
John Doe
11.1k11238
edited Jan 12 at 21:38
John Doe
11.1k11238
11.1k11238
answered Jan 12 at 19:27
Peter SzilasPeter Szilas
11.2k2822
answered Jan 12 at 19:27
Peter SzilasPeter Szilas
11.2k2822
answered Jan 12 at 19:27
Peter SzilasPeter Szilas
11.2k2822
11.2k2822
$begingroup$
Joe Doe.Thanks:)
$endgroup$
– Peter Szilas
Jan 12 at 22:00
add a comment |
$begingroup$
Joe Doe.Thanks:)
$endgroup$
– Peter Szilas
Jan 12 at 22:00
$begingroup$
Joe Doe.Thanks:)
$endgroup$
– Peter Szilas
Jan 12 at 22:00
$begingroup$
Joe Doe.Thanks:)
$endgroup$
– Peter Szilas
Jan 12 at 22:00
add a comment |
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$begingroup$
Also: $frac{1 + x}{1 + 2x} = 1 + frac{1}{frac{1 + 2x}{-x}}$
$endgroup$
– user626177
Jan 16 at 1:59