Continuity of a Piecewise Function $ - e^-1/x $ on $[0,1]$
Clash Royale CLAN TAG#URR8PPP
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1
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Prove the following function is continuous on $[0,1]$:
$$f(x) =
begincases
mathrme^-frac1x & textif & x not=0 \
0 & textif & x=0
endcases
$$
My attempt at a proof:
Proof. Consider two cases: showing $f$ is continuous at (1) $x not= 0$, and (2) $x = 0$.
(1) If $x not=0$, take $g(x)= -frac1x$, and $q(x) = mathrme^x$. Then, on $(0,1]$, we have that $f(x) = (q circ g)(x)$. Both $q$, and $g$ are continuous on $(0,1]$, so $f$ is continuous on $(0,1]$.*
(2) If $x = 0$, we must show $$lim_x to 0 f(x) = f(0) = 0$$
Notice that $f(x) = mathrme^-frac1x$ for all $x not= 0$. Thus, $$lim_x to 0 f(x) = lim_x to 0 mathrme^-frac1x$$
This is as far as I got - I don't know how to continue. Is this the right approach? If not, how should I approach the problem?
* Take for granted that $-frac1x$ is continuous on $(0,1]$.
calculus real-analysis analysis
edited Jun 27 at 2:03
HK Lee
13.5k41857
asked Oct 4 '16 at 5:57
user375004
62
add a comment |Â
up vote
1
down vote
favorite
Prove the following function is continuous on $[0,1]$:
$$f(x) =
begincases
mathrme^-frac1x & textif & x not=0 \
0 & textif & x=0
endcases
$$
My attempt at a proof:
Proof. Consider two cases: showing $f$ is continuous at (1) $x not= 0$, and (2) $x = 0$.
(1) If $x not=0$, take $g(x)= -frac1x$, and $q(x) = mathrme^x$. Then, on $(0,1]$, we have that $f(x) = (q circ g)(x)$. Both $q$, and $g$ are continuous on $(0,1]$, so $f$ is continuous on $(0,1]$.*
(2) If $x = 0$, we must show $$lim_x to 0 f(x) = f(0) = 0$$
Notice that $f(x) = mathrme^-frac1x$ for all $x not= 0$. Thus, $$lim_x to 0 f(x) = lim_x to 0 mathrme^-frac1x$$
This is as far as I got - I don't know how to continue. Is this the right approach? If not, how should I approach the problem?
* Take for granted that $-frac1x$ is continuous on $(0,1]$.
calculus real-analysis analysis
edited Jun 27 at 2:03
HK Lee
13.5k41857
asked Oct 4 '16 at 5:57
user375004
62
1
Since $x$ approaches 0 from the right side, i.e. $x$ is positive: $$lim_xto 0-frac1x=-infty$$ you may continue from here
– polfosol
Oct 4 '16 at 7:32
And not that $$lim_xto x_0 fcirc g(x)=fleft(lim_xto x_0g(x)right)$$
– polfosol
Oct 4 '16 at 7:35
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Prove the following function is continuous on $[0,1]$:
$$f(x) =
begincases
mathrme^-frac1x & textif & x not=0 \
0 & textif & x=0
endcases
$$
My attempt at a proof:
Proof. Consider two cases: showing $f$ is continuous at (1) $x not= 0$, and (2) $x = 0$.
(1) If $x not=0$, take $g(x)= -frac1x$, and $q(x) = mathrme^x$. Then, on $(0,1]$, we have that $f(x) = (q circ g)(x)$. Both $q$, and $g$ are continuous on $(0,1]$, so $f$ is continuous on $(0,1]$.*
(2) If $x = 0$, we must show $$lim_x to 0 f(x) = f(0) = 0$$
Notice that $f(x) = mathrme^-frac1x$ for all $x not= 0$. Thus, $$lim_x to 0 f(x) = lim_x to 0 mathrme^-frac1x$$
This is as far as I got - I don't know how to continue. Is this the right approach? If not, how should I approach the problem?
* Take for granted that $-frac1x$ is continuous on $(0,1]$.
calculus real-analysis analysis
edited Jun 27 at 2:03
HK Lee
13.5k41857
asked Oct 4 '16 at 5:57
user375004
62
Prove the following function is continuous on $[0,1]$:
$$f(x) =
begincases
mathrme^-frac1x & textif & x not=0 \
0 & textif & x=0
endcases
$$
My attempt at a proof:
Proof. Consider two cases: showing $f$ is continuous at (1) $x not= 0$, and (2) $x = 0$.
(1) If $x not=0$, take $g(x)= -frac1x$, and $q(x) = mathrme^x$. Then, on $(0,1]$, we have that $f(x) = (q circ g)(x)$. Both $q$, and $g$ are continuous on $(0,1]$, so $f$ is continuous on $(0,1]$.*
(2) If $x = 0$, we must show $$lim_x to 0 f(x) = f(0) = 0$$
Notice that $f(x) = mathrme^-frac1x$ for all $x not= 0$. Thus, $$lim_x to 0 f(x) = lim_x to 0 mathrme^-frac1x$$
This is as far as I got - I don't know how to continue. Is this the right approach? If not, how should I approach the problem?
* Take for granted that $-frac1x$ is continuous on $(0,1]$.
calculus real-analysis analysis
calculus real-analysis analysis
edited Jun 27 at 2:03
HK Lee
13.5k41857
asked Oct 4 '16 at 5:57
user375004
62
edited Jun 27 at 2:03
HK Lee
13.5k41857
asked Oct 4 '16 at 5:57
user375004
62
edited Jun 27 at 2:03
HK Lee
13.5k41857
edited Jun 27 at 2:03
HK Lee
13.5k41857
edited Jun 27 at 2:03
HK Lee
13.5k41857
13.5k41857
asked Oct 4 '16 at 5:57
user375004
62
asked Oct 4 '16 at 5:57
user375004
62
asked Oct 4 '16 at 5:57
user375004
62
62
1
Since $x$ approaches 0 from the right side, i.e. $x$ is positive: $$lim_xto 0-frac1x=-infty$$ you may continue from here
– polfosol
Oct 4 '16 at 7:32
And not that $$lim_xto x_0 fcirc g(x)=fleft(lim_xto x_0g(x)right)$$
– polfosol
Oct 4 '16 at 7:35
add a comment |Â
1
Since $x$ approaches 0 from the right side, i.e. $x$ is positive: $$lim_xto 0-frac1x=-infty$$ you may continue from here
– polfosol
Oct 4 '16 at 7:32
And not that $$lim_xto x_0 fcirc g(x)=fleft(lim_xto x_0g(x)right)$$
– polfosol
Oct 4 '16 at 7:35
1
1
Since $x$ approaches 0 from the right side, i.e. $x$ is positive: $$lim_xto 0-frac1x=-infty$$ you may continue from here
– polfosol
Oct 4 '16 at 7:32
Since $x$ approaches 0 from the right side, i.e. $x$ is positive: $$lim_xto 0-frac1x=-infty$$ you may continue from here
– polfosol
Oct 4 '16 at 7:32
And not that $$lim_xto x_0 fcirc g(x)=fleft(lim_xto x_0g(x)right)$$
– polfosol
Oct 4 '16 at 7:35
And not that $$lim_xto x_0 fcirc g(x)=fleft(lim_xto x_0g(x)right)$$
– polfosol
Oct 4 '16 at 7:35
add a comment |Â
1 Answer
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Let $u=1/x$ so that $urightarrowinfty$ as $xrightarrow 0^+$.
Then $limlimits_xrightarrow 0^+f(x) = limlimits_xrightarrow 0^+e^-1/x = limlimits_urightarrowinftye^-u = limlimits_urightarrowinftyfrac1e^u = 0$
answered Feb 6 '17 at 3:28
Tanner Strunk
613315
Found this question while trying to figure out how to prove that the same function is $C^infty$, i.e. that it has partial derivatives of all orders...
– Tanner Strunk
Feb 6 '17 at 3:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $u=1/x$ so that $urightarrowinfty$ as $xrightarrow 0^+$.
Then $limlimits_xrightarrow 0^+f(x) = limlimits_xrightarrow 0^+e^-1/x = limlimits_urightarrowinftye^-u = limlimits_urightarrowinftyfrac1e^u = 0$
answered Feb 6 '17 at 3:28
Tanner Strunk
613315
Found this question while trying to figure out how to prove that the same function is $C^infty$, i.e. that it has partial derivatives of all orders...
– Tanner Strunk
Feb 6 '17 at 3:30
add a comment |Â
up vote
0
down vote
Let $u=1/x$ so that $urightarrowinfty$ as $xrightarrow 0^+$.
Then $limlimits_xrightarrow 0^+f(x) = limlimits_xrightarrow 0^+e^-1/x = limlimits_urightarrowinftye^-u = limlimits_urightarrowinftyfrac1e^u = 0$
answered Feb 6 '17 at 3:28
Tanner Strunk
613315
Found this question while trying to figure out how to prove that the same function is $C^infty$, i.e. that it has partial derivatives of all orders...
– Tanner Strunk
Feb 6 '17 at 3:30
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $u=1/x$ so that $urightarrowinfty$ as $xrightarrow 0^+$.
Then $limlimits_xrightarrow 0^+f(x) = limlimits_xrightarrow 0^+e^-1/x = limlimits_urightarrowinftye^-u = limlimits_urightarrowinftyfrac1e^u = 0$
answered Feb 6 '17 at 3:28
Tanner Strunk
613315
Let $u=1/x$ so that $urightarrowinfty$ as $xrightarrow 0^+$.
Then $limlimits_xrightarrow 0^+f(x) = limlimits_xrightarrow 0^+e^-1/x = limlimits_urightarrowinftye^-u = limlimits_urightarrowinftyfrac1e^u = 0$
answered Feb 6 '17 at 3:28
Tanner Strunk
613315
answered Feb 6 '17 at 3:28
Tanner Strunk
613315
answered Feb 6 '17 at 3:28
Tanner Strunk
613315
answered Feb 6 '17 at 3:28
Tanner Strunk
613315
613315
Found this question while trying to figure out how to prove that the same function is $C^infty$, i.e. that it has partial derivatives of all orders...
– Tanner Strunk
Feb 6 '17 at 3:30
add a comment |Â
Found this question while trying to figure out how to prove that the same function is $C^infty$, i.e. that it has partial derivatives of all orders...
– Tanner Strunk
Feb 6 '17 at 3:30
Found this question while trying to figure out how to prove that the same function is $C^infty$, i.e. that it has partial derivatives of all orders...
– Tanner Strunk
Feb 6 '17 at 3:30
Found this question while trying to figure out how to prove that the same function is $C^infty$, i.e. that it has partial derivatives of all orders...
– Tanner Strunk
Feb 6 '17 at 3:30
add a comment |Â
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1
Since $x$ approaches 0 from the right side, i.e. $x$ is positive: $$lim_xto 0-frac1x=-infty$$ you may continue from here
– polfosol
Oct 4 '16 at 7:32
And not that $$lim_xto x_0 fcirc g(x)=fleft(lim_xto x_0g(x)right)$$
– polfosol
Oct 4 '16 at 7:35