Limit using definition of $e$
$begingroup$
I'm trying to calculate the following limit:
$$
lim_{xto0}bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x}
$$
My instinct is as follows: use the binomial theorem on the square root, cancel higher order terms (i.e treat the square root as $1$), then appeal to the limit definition of $e$ to get the answer $exp(-4b)$. However, this does not feel terribly rigorous and I'm unsure if it's right.
Does anyone have any advice?
real-analysis
edited Jan 7 at 21:19
egreg
185k1486206
asked Jan 7 at 21:14
Bob Garth Bob Garth
61
$endgroup$
add a comment |
$begingroup$
I'm trying to calculate the following limit:
$$
lim_{xto0}bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x}
$$
My instinct is as follows: use the binomial theorem on the square root, cancel higher order terms (i.e treat the square root as $1$), then appeal to the limit definition of $e$ to get the answer $exp(-4b)$. However, this does not feel terribly rigorous and I'm unsure if it's right.
Does anyone have any advice?
real-analysis
edited Jan 7 at 21:19
egreg
185k1486206
asked Jan 7 at 21:14
Bob Garth Bob Garth
61
$endgroup$
add a comment |
$begingroup$
I'm trying to calculate the following limit:
$$
lim_{xto0}bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x}
$$
My instinct is as follows: use the binomial theorem on the square root, cancel higher order terms (i.e treat the square root as $1$), then appeal to the limit definition of $e$ to get the answer $exp(-4b)$. However, this does not feel terribly rigorous and I'm unsure if it's right.
Does anyone have any advice?
real-analysis
edited Jan 7 at 21:19
egreg
185k1486206
asked Jan 7 at 21:14
Bob Garth Bob Garth
61
$endgroup$
I'm trying to calculate the following limit:
$$
lim_{xto0}bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x}
$$
My instinct is as follows: use the binomial theorem on the square root, cancel higher order terms (i.e treat the square root as $1$), then appeal to the limit definition of $e$ to get the answer $exp(-4b)$. However, this does not feel terribly rigorous and I'm unsure if it's right.
Does anyone have any advice?
real-analysis
real-analysis
edited Jan 7 at 21:19
egreg
185k1486206
asked Jan 7 at 21:14
Bob Garth Bob Garth
61
edited Jan 7 at 21:19
egreg
185k1486206
asked Jan 7 at 21:14
Bob Garth Bob Garth
61
edited Jan 7 at 21:19
egreg
185k1486206
edited Jan 7 at 21:19
egreg
185k1486206
edited Jan 7 at 21:19
egreg
185k1486206
185k1486206
asked Jan 7 at 21:14
Bob Garth Bob Garth
61
asked Jan 7 at 21:14
Bob Garth Bob Garth
61
asked Jan 7 at 21:14
Bob Garth Bob Garth
61
61
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
You can use little-$o$ notation. You're taking the limit of $$(1-4x+8x^2+o(x^2))^{b/x}=exp frac{b}{x}(-4x+o(x))=exp (-4b+o(1))=exp(-4b)+o(1).$$
answered Jan 7 at 21:25
J.G.J.G.
32.3k23250
$endgroup$
$begingroup$
@egreg Thanks; fixed.
$endgroup$
– J.G.
Jan 7 at 21:35
add a comment |
$begingroup$
Letting $L = lim_{X to 0} (-4x+sqrt{16x^2+1})^{frac{b}{x}}$ we have;
$$ln{L}=b cdot lim_{X to 0} frac{ln{-4x+sqrt{16x^2+1}}}{x} $$
By using L'Hôpital's Rule, this is equal to:
$$ln{L}=b cdot lim_{X to 0} frac{-4+frac{32x}{2 cdot sqrt{16x^2+1}}}{-4x+sqrt{16x^2+1}} $$
$$=-4 cdot b$$
$$therefore L=e^{-4 cdot b}$$
answered Jan 7 at 21:26
Peter ForemanPeter Foreman
4,5401216
$endgroup$
$begingroup$
uselimlimits_{xto 0}for $limlimits_{xto 0}$
$endgroup$
– zwim
Jan 7 at 21:27
$begingroup$
A few judicious parentheses might help here.
$endgroup$
– TonyK
Jan 7 at 21:30
add a comment |
$begingroup$
It's basically right:
$$
-4x+sqrt{1+16x^2}=-4x+1+frac{16x^2}{2}+o(x^2)=1-4x+o(x)
$$
Now you can substitute $-4x=y$, so the limit becomes
$$
lim_{yto0}bigl((1+y+o(y))^{1/y}bigr)^{-4b}
$$
and $lim_{yto0}(1+y)^{1/y}=e$ is known.
answered Jan 7 at 21:39
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
Consider $$y=bigr( sqrt{16x^2 + 1}-4xbigr)^{frac bx}$$ By Taylor or binomial expansion, we have
$$sqrt{16x^2 + 1}=1+8 x^2+Oleft(x^4right)$$ that is to say
$$sqrt{16x^2 + 1}-4x=1-4 x+8 x^2+Oleft(x^4right)$$ Now
$$log(y)=frac bx logbigr( sqrt{16x^2 + 1}-4xbigr)=frac bxlogbigr(1-4 x+8 x^2+Oleft(x^4right)bigr)=frac bxbigr( -4 x+frac{32 x^3}{3}+Oleft(x^4right)bigr)$$
$$log(y)=-4 b+frac{32 b x^2}{3}+Oleft(x^3right)$$ Continue with Taylor
$$y=e^{log(y)}=e^{-4 b}+frac{32}{3} b e^{-4 b} x^2+Oleft(x^3right)$$ which shows the limit and how it is approached.
answered Jan 8 at 6:42
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Here is a way by "enforcing" a $color{red}{1}$ in the brackets and using the standard limit for $e$:
$y_n stackrel{n to infty}{longrightarrow}0 Rightarrow (1+y_n)^{frac{1}{y_n}}stackrel{n to infty}{longrightarrow}e$
begin{eqnarray*}
bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x} & = & left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{b/x}\
& = & left(underbrace{left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{frac{1}{color{blue}{-4x + sqrt{16x^2+1} -1}}}}_{stackrel{xto 0}{rightarrow}e} right)^{bcdot underbrace{frac{color{blue}{-4x + sqrt{16x^2+1} -1}}{x}}_{stackrel{xto 0}{rightarrow}-4=f'(0),; f(x) = -4x + sqrt{16x^2+1}}} \
& stackrel{xto 0}{longrightarrow} & e^{bcdot (-4)}
end{eqnarray*}
answered Jan 8 at 12:11
trancelocationtrancelocation
13.3k1827
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use little-$o$ notation. You're taking the limit of $$(1-4x+8x^2+o(x^2))^{b/x}=exp frac{b}{x}(-4x+o(x))=exp (-4b+o(1))=exp(-4b)+o(1).$$
answered Jan 7 at 21:25
J.G.J.G.
32.3k23250
$endgroup$
$begingroup$
@egreg Thanks; fixed.
$endgroup$
– J.G.
Jan 7 at 21:35
add a comment |
$begingroup$
You can use little-$o$ notation. You're taking the limit of $$(1-4x+8x^2+o(x^2))^{b/x}=exp frac{b}{x}(-4x+o(x))=exp (-4b+o(1))=exp(-4b)+o(1).$$
answered Jan 7 at 21:25
J.G.J.G.
32.3k23250
$endgroup$
$begingroup$
@egreg Thanks; fixed.
$endgroup$
– J.G.
Jan 7 at 21:35
add a comment |
$begingroup$
You can use little-$o$ notation. You're taking the limit of $$(1-4x+8x^2+o(x^2))^{b/x}=exp frac{b}{x}(-4x+o(x))=exp (-4b+o(1))=exp(-4b)+o(1).$$
answered Jan 7 at 21:25
J.G.J.G.
32.3k23250
$endgroup$
You can use little-$o$ notation. You're taking the limit of $$(1-4x+8x^2+o(x^2))^{b/x}=exp frac{b}{x}(-4x+o(x))=exp (-4b+o(1))=exp(-4b)+o(1).$$
answered Jan 7 at 21:25
J.G.J.G.
32.3k23250
edited Jan 7 at 21:35
answered Jan 7 at 21:25
J.G.J.G.
32.3k23250
answered Jan 7 at 21:25
J.G.J.G.
32.3k23250
answered Jan 7 at 21:25
J.G.J.G.
32.3k23250
32.3k23250
$begingroup$
@egreg Thanks; fixed.
$endgroup$
– J.G.
Jan 7 at 21:35
add a comment |
$begingroup$
@egreg Thanks; fixed.
$endgroup$
– J.G.
Jan 7 at 21:35
$begingroup$
@egreg Thanks; fixed.
$endgroup$
– J.G.
Jan 7 at 21:35
$begingroup$
@egreg Thanks; fixed.
$endgroup$
– J.G.
Jan 7 at 21:35
add a comment |
$begingroup$
Letting $L = lim_{X to 0} (-4x+sqrt{16x^2+1})^{frac{b}{x}}$ we have;
$$ln{L}=b cdot lim_{X to 0} frac{ln{-4x+sqrt{16x^2+1}}}{x} $$
By using L'Hôpital's Rule, this is equal to:
$$ln{L}=b cdot lim_{X to 0} frac{-4+frac{32x}{2 cdot sqrt{16x^2+1}}}{-4x+sqrt{16x^2+1}} $$
$$=-4 cdot b$$
$$therefore L=e^{-4 cdot b}$$
answered Jan 7 at 21:26
Peter ForemanPeter Foreman
4,5401216
$endgroup$
$begingroup$
uselimlimits_{xto 0}for $limlimits_{xto 0}$
$endgroup$
– zwim
Jan 7 at 21:27
$begingroup$
A few judicious parentheses might help here.
$endgroup$
– TonyK
Jan 7 at 21:30
add a comment |
$begingroup$
Letting $L = lim_{X to 0} (-4x+sqrt{16x^2+1})^{frac{b}{x}}$ we have;
$$ln{L}=b cdot lim_{X to 0} frac{ln{-4x+sqrt{16x^2+1}}}{x} $$
By using L'Hôpital's Rule, this is equal to:
$$ln{L}=b cdot lim_{X to 0} frac{-4+frac{32x}{2 cdot sqrt{16x^2+1}}}{-4x+sqrt{16x^2+1}} $$
$$=-4 cdot b$$
$$therefore L=e^{-4 cdot b}$$
answered Jan 7 at 21:26
Peter ForemanPeter Foreman
4,5401216
$endgroup$
$begingroup$
uselimlimits_{xto 0}for $limlimits_{xto 0}$
$endgroup$
– zwim
Jan 7 at 21:27
$begingroup$
A few judicious parentheses might help here.
$endgroup$
– TonyK
Jan 7 at 21:30
add a comment |
$begingroup$
Letting $L = lim_{X to 0} (-4x+sqrt{16x^2+1})^{frac{b}{x}}$ we have;
$$ln{L}=b cdot lim_{X to 0} frac{ln{-4x+sqrt{16x^2+1}}}{x} $$
By using L'Hôpital's Rule, this is equal to:
$$ln{L}=b cdot lim_{X to 0} frac{-4+frac{32x}{2 cdot sqrt{16x^2+1}}}{-4x+sqrt{16x^2+1}} $$
$$=-4 cdot b$$
$$therefore L=e^{-4 cdot b}$$
answered Jan 7 at 21:26
Peter ForemanPeter Foreman
4,5401216
$endgroup$
Letting $L = lim_{X to 0} (-4x+sqrt{16x^2+1})^{frac{b}{x}}$ we have;
$$ln{L}=b cdot lim_{X to 0} frac{ln{-4x+sqrt{16x^2+1}}}{x} $$
By using L'Hôpital's Rule, this is equal to:
$$ln{L}=b cdot lim_{X to 0} frac{-4+frac{32x}{2 cdot sqrt{16x^2+1}}}{-4x+sqrt{16x^2+1}} $$
$$=-4 cdot b$$
$$therefore L=e^{-4 cdot b}$$
answered Jan 7 at 21:26
Peter ForemanPeter Foreman
4,5401216
answered Jan 7 at 21:26
Peter ForemanPeter Foreman
4,5401216
answered Jan 7 at 21:26
Peter ForemanPeter Foreman
4,5401216
answered Jan 7 at 21:26
Peter ForemanPeter Foreman
4,5401216
4,5401216
$begingroup$
uselimlimits_{xto 0}for $limlimits_{xto 0}$
$endgroup$
– zwim
Jan 7 at 21:27
$begingroup$
A few judicious parentheses might help here.
$endgroup$
– TonyK
Jan 7 at 21:30
add a comment |
$begingroup$
uselimlimits_{xto 0}for $limlimits_{xto 0}$
$endgroup$
– zwim
Jan 7 at 21:27
$begingroup$
A few judicious parentheses might help here.
$endgroup$
– TonyK
Jan 7 at 21:30
$begingroup$
use
limlimits_{xto 0} for $limlimits_{xto 0}$$endgroup$
– zwim
Jan 7 at 21:27
$begingroup$
use
limlimits_{xto 0} for $limlimits_{xto 0}$$endgroup$
– zwim
Jan 7 at 21:27
$begingroup$
A few judicious parentheses might help here.
$endgroup$
– TonyK
Jan 7 at 21:30
$begingroup$
A few judicious parentheses might help here.
$endgroup$
– TonyK
Jan 7 at 21:30
add a comment |
$begingroup$
It's basically right:
$$
-4x+sqrt{1+16x^2}=-4x+1+frac{16x^2}{2}+o(x^2)=1-4x+o(x)
$$
Now you can substitute $-4x=y$, so the limit becomes
$$
lim_{yto0}bigl((1+y+o(y))^{1/y}bigr)^{-4b}
$$
and $lim_{yto0}(1+y)^{1/y}=e$ is known.
answered Jan 7 at 21:39
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
It's basically right:
$$
-4x+sqrt{1+16x^2}=-4x+1+frac{16x^2}{2}+o(x^2)=1-4x+o(x)
$$
Now you can substitute $-4x=y$, so the limit becomes
$$
lim_{yto0}bigl((1+y+o(y))^{1/y}bigr)^{-4b}
$$
and $lim_{yto0}(1+y)^{1/y}=e$ is known.
answered Jan 7 at 21:39
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
It's basically right:
$$
-4x+sqrt{1+16x^2}=-4x+1+frac{16x^2}{2}+o(x^2)=1-4x+o(x)
$$
Now you can substitute $-4x=y$, so the limit becomes
$$
lim_{yto0}bigl((1+y+o(y))^{1/y}bigr)^{-4b}
$$
and $lim_{yto0}(1+y)^{1/y}=e$ is known.
answered Jan 7 at 21:39
egregegreg
185k1486206
$endgroup$
It's basically right:
$$
-4x+sqrt{1+16x^2}=-4x+1+frac{16x^2}{2}+o(x^2)=1-4x+o(x)
$$
Now you can substitute $-4x=y$, so the limit becomes
$$
lim_{yto0}bigl((1+y+o(y))^{1/y}bigr)^{-4b}
$$
and $lim_{yto0}(1+y)^{1/y}=e$ is known.
answered Jan 7 at 21:39
egregegreg
185k1486206
answered Jan 7 at 21:39
egregegreg
185k1486206
answered Jan 7 at 21:39
egregegreg
185k1486206
answered Jan 7 at 21:39
egregegreg
185k1486206
185k1486206
add a comment |
add a comment |
$begingroup$
Consider $$y=bigr( sqrt{16x^2 + 1}-4xbigr)^{frac bx}$$ By Taylor or binomial expansion, we have
$$sqrt{16x^2 + 1}=1+8 x^2+Oleft(x^4right)$$ that is to say
$$sqrt{16x^2 + 1}-4x=1-4 x+8 x^2+Oleft(x^4right)$$ Now
$$log(y)=frac bx logbigr( sqrt{16x^2 + 1}-4xbigr)=frac bxlogbigr(1-4 x+8 x^2+Oleft(x^4right)bigr)=frac bxbigr( -4 x+frac{32 x^3}{3}+Oleft(x^4right)bigr)$$
$$log(y)=-4 b+frac{32 b x^2}{3}+Oleft(x^3right)$$ Continue with Taylor
$$y=e^{log(y)}=e^{-4 b}+frac{32}{3} b e^{-4 b} x^2+Oleft(x^3right)$$ which shows the limit and how it is approached.
answered Jan 8 at 6:42
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Consider $$y=bigr( sqrt{16x^2 + 1}-4xbigr)^{frac bx}$$ By Taylor or binomial expansion, we have
$$sqrt{16x^2 + 1}=1+8 x^2+Oleft(x^4right)$$ that is to say
$$sqrt{16x^2 + 1}-4x=1-4 x+8 x^2+Oleft(x^4right)$$ Now
$$log(y)=frac bx logbigr( sqrt{16x^2 + 1}-4xbigr)=frac bxlogbigr(1-4 x+8 x^2+Oleft(x^4right)bigr)=frac bxbigr( -4 x+frac{32 x^3}{3}+Oleft(x^4right)bigr)$$
$$log(y)=-4 b+frac{32 b x^2}{3}+Oleft(x^3right)$$ Continue with Taylor
$$y=e^{log(y)}=e^{-4 b}+frac{32}{3} b e^{-4 b} x^2+Oleft(x^3right)$$ which shows the limit and how it is approached.
answered Jan 8 at 6:42
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Consider $$y=bigr( sqrt{16x^2 + 1}-4xbigr)^{frac bx}$$ By Taylor or binomial expansion, we have
$$sqrt{16x^2 + 1}=1+8 x^2+Oleft(x^4right)$$ that is to say
$$sqrt{16x^2 + 1}-4x=1-4 x+8 x^2+Oleft(x^4right)$$ Now
$$log(y)=frac bx logbigr( sqrt{16x^2 + 1}-4xbigr)=frac bxlogbigr(1-4 x+8 x^2+Oleft(x^4right)bigr)=frac bxbigr( -4 x+frac{32 x^3}{3}+Oleft(x^4right)bigr)$$
$$log(y)=-4 b+frac{32 b x^2}{3}+Oleft(x^3right)$$ Continue with Taylor
$$y=e^{log(y)}=e^{-4 b}+frac{32}{3} b e^{-4 b} x^2+Oleft(x^3right)$$ which shows the limit and how it is approached.
answered Jan 8 at 6:42
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
Consider $$y=bigr( sqrt{16x^2 + 1}-4xbigr)^{frac bx}$$ By Taylor or binomial expansion, we have
$$sqrt{16x^2 + 1}=1+8 x^2+Oleft(x^4right)$$ that is to say
$$sqrt{16x^2 + 1}-4x=1-4 x+8 x^2+Oleft(x^4right)$$ Now
$$log(y)=frac bx logbigr( sqrt{16x^2 + 1}-4xbigr)=frac bxlogbigr(1-4 x+8 x^2+Oleft(x^4right)bigr)=frac bxbigr( -4 x+frac{32 x^3}{3}+Oleft(x^4right)bigr)$$
$$log(y)=-4 b+frac{32 b x^2}{3}+Oleft(x^3right)$$ Continue with Taylor
$$y=e^{log(y)}=e^{-4 b}+frac{32}{3} b e^{-4 b} x^2+Oleft(x^3right)$$ which shows the limit and how it is approached.
answered Jan 8 at 6:42
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 8 at 6:42
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 8 at 6:42
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 8 at 6:42
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
$begingroup$
Here is a way by "enforcing" a $color{red}{1}$ in the brackets and using the standard limit for $e$:
$y_n stackrel{n to infty}{longrightarrow}0 Rightarrow (1+y_n)^{frac{1}{y_n}}stackrel{n to infty}{longrightarrow}e$
begin{eqnarray*}
bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x} & = & left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{b/x}\
& = & left(underbrace{left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{frac{1}{color{blue}{-4x + sqrt{16x^2+1} -1}}}}_{stackrel{xto 0}{rightarrow}e} right)^{bcdot underbrace{frac{color{blue}{-4x + sqrt{16x^2+1} -1}}{x}}_{stackrel{xto 0}{rightarrow}-4=f'(0),; f(x) = -4x + sqrt{16x^2+1}}} \
& stackrel{xto 0}{longrightarrow} & e^{bcdot (-4)}
end{eqnarray*}
answered Jan 8 at 12:11
trancelocationtrancelocation
13.3k1827
$endgroup$
add a comment |
$begingroup$
Here is a way by "enforcing" a $color{red}{1}$ in the brackets and using the standard limit for $e$:
$y_n stackrel{n to infty}{longrightarrow}0 Rightarrow (1+y_n)^{frac{1}{y_n}}stackrel{n to infty}{longrightarrow}e$
begin{eqnarray*}
bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x} & = & left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{b/x}\
& = & left(underbrace{left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{frac{1}{color{blue}{-4x + sqrt{16x^2+1} -1}}}}_{stackrel{xto 0}{rightarrow}e} right)^{bcdot underbrace{frac{color{blue}{-4x + sqrt{16x^2+1} -1}}{x}}_{stackrel{xto 0}{rightarrow}-4=f'(0),; f(x) = -4x + sqrt{16x^2+1}}} \
& stackrel{xto 0}{longrightarrow} & e^{bcdot (-4)}
end{eqnarray*}
answered Jan 8 at 12:11
trancelocationtrancelocation
13.3k1827
$endgroup$
add a comment |
$begingroup$
Here is a way by "enforcing" a $color{red}{1}$ in the brackets and using the standard limit for $e$:
$y_n stackrel{n to infty}{longrightarrow}0 Rightarrow (1+y_n)^{frac{1}{y_n}}stackrel{n to infty}{longrightarrow}e$
begin{eqnarray*}
bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x} & = & left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{b/x}\
& = & left(underbrace{left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{frac{1}{color{blue}{-4x + sqrt{16x^2+1} -1}}}}_{stackrel{xto 0}{rightarrow}e} right)^{bcdot underbrace{frac{color{blue}{-4x + sqrt{16x^2+1} -1}}{x}}_{stackrel{xto 0}{rightarrow}-4=f'(0),; f(x) = -4x + sqrt{16x^2+1}}} \
& stackrel{xto 0}{longrightarrow} & e^{bcdot (-4)}
end{eqnarray*}
answered Jan 8 at 12:11
trancelocationtrancelocation
13.3k1827
$endgroup$
Here is a way by "enforcing" a $color{red}{1}$ in the brackets and using the standard limit for $e$:
$y_n stackrel{n to infty}{longrightarrow}0 Rightarrow (1+y_n)^{frac{1}{y_n}}stackrel{n to infty}{longrightarrow}e$
begin{eqnarray*}
bigl(-4x + sqrt{16x^2 + 1},bigr)^{b/x} & = & left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{b/x}\
& = & left(underbrace{left(color{red}{1} + left(color{blue}{-4x + sqrt{16x^2+1} -1}right) right)^{frac{1}{color{blue}{-4x + sqrt{16x^2+1} -1}}}}_{stackrel{xto 0}{rightarrow}e} right)^{bcdot underbrace{frac{color{blue}{-4x + sqrt{16x^2+1} -1}}{x}}_{stackrel{xto 0}{rightarrow}-4=f'(0),; f(x) = -4x + sqrt{16x^2+1}}} \
& stackrel{xto 0}{longrightarrow} & e^{bcdot (-4)}
end{eqnarray*}
answered Jan 8 at 12:11
trancelocationtrancelocation
13.3k1827
answered Jan 8 at 12:11
trancelocationtrancelocation
13.3k1827
answered Jan 8 at 12:11
trancelocationtrancelocation
13.3k1827
answered Jan 8 at 12:11
trancelocationtrancelocation
13.3k1827
13.3k1827
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