Mixing
Positivity of a limit
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Let $f:mathbb R^2 rightarrow mathbb R$. Show that if $f(x)>0$ for all $x$ and $lim_{xrightarrow a}f(x)$ exists, then $lim_{xrightarrow 0}f(x)geq0$.
My attempt:
Let $lim_{xrightarrow 0}f(x)=l$. By definition this means that for all $epsilon>0 $ there is $delta>0$ such that $|x-0|<delta$ implies $|f(x)-l|<epsilon.$ Hence, $f(x)-epsilon <l<f(x)+epsilon$.
Can we make $epsilon$ very small, so that $epsilon<f(x)$ (then 0<$f(x)-epsilon <l$), which is what we wanted to prove?
real-analysis
asked Feb 3 at 0:06
dxdydzdxdydz
49110
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb R^2 rightarrow mathbb R$. Show that if $f(x)>0$ for all $x$ and $lim_{xrightarrow a}f(x)$ exists, then $lim_{xrightarrow 0}f(x)geq0$.
My attempt:
Let $lim_{xrightarrow 0}f(x)=l$. By definition this means that for all $epsilon>0 $ there is $delta>0$ such that $|x-0|<delta$ implies $|f(x)-l|<epsilon.$ Hence, $f(x)-epsilon <l<f(x)+epsilon$.
Can we make $epsilon$ very small, so that $epsilon<f(x)$ (then 0<$f(x)-epsilon <l$), which is what we wanted to prove?
real-analysis
asked Feb 3 at 0:06
dxdydzdxdydz
49110
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb R^2 rightarrow mathbb R$. Show that if $f(x)>0$ for all $x$ and $lim_{xrightarrow a}f(x)$ exists, then $lim_{xrightarrow 0}f(x)geq0$.
My attempt:
Let $lim_{xrightarrow 0}f(x)=l$. By definition this means that for all $epsilon>0 $ there is $delta>0$ such that $|x-0|<delta$ implies $|f(x)-l|<epsilon.$ Hence, $f(x)-epsilon <l<f(x)+epsilon$.
Can we make $epsilon$ very small, so that $epsilon<f(x)$ (then 0<$f(x)-epsilon <l$), which is what we wanted to prove?
real-analysis
asked Feb 3 at 0:06
dxdydzdxdydz
49110
$endgroup$
Let $f:mathbb R^2 rightarrow mathbb R$. Show that if $f(x)>0$ for all $x$ and $lim_{xrightarrow a}f(x)$ exists, then $lim_{xrightarrow 0}f(x)geq0$.
My attempt:
Let $lim_{xrightarrow 0}f(x)=l$. By definition this means that for all $epsilon>0 $ there is $delta>0$ such that $|x-0|<delta$ implies $|f(x)-l|<epsilon.$ Hence, $f(x)-epsilon <l<f(x)+epsilon$.
Can we make $epsilon$ very small, so that $epsilon<f(x)$ (then 0<$f(x)-epsilon <l$), which is what we wanted to prove?
real-analysis
real-analysis
asked Feb 3 at 0:06
dxdydzdxdydz
49110
asked Feb 3 at 0:06
dxdydzdxdydz
49110
asked Feb 3 at 0:06
dxdydzdxdydz
49110
asked Feb 3 at 0:06
dxdydzdxdydz
49110
asked Feb 3 at 0:06
dxdydzdxdydz
49110
49110
add a comment |
add a comment |
2 Answers
2
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Prove by contradiction. Suppose $L=lim_{x to a} f(x) <0$. Let $0<epsilon <-L$. Then, for $|x-a|$ sufficiently small we get $|f(x)-L| <epsilon$ which implies $f(x) <L+epsilon <0$, a contradiction.
answered Feb 3 at 0:48
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
$endgroup$
add a comment |
$begingroup$
Since $f(x)>0$, $-epsilon<f(x)-epsilon$. You have shown that for every $epsilon>0$, there exists $x$ such that $-epsilon<f(x)-epsilonleq l$ implies $lgeq 0$.
answered Feb 3 at 0:16
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
$begingroup$
but since $epsilon>0$, then $-epsilon<0$, so $𝑓(𝑥)−𝜖$ can be negative too, no?
$endgroup$
– dxdydz
Feb 3 at 0:24
$begingroup$
A number which is superior to every negative number is positive, if not, $l<0, l<{lover 2}<0$.
$endgroup$
– Tsemo Aristide
Feb 3 at 0:28
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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$begingroup$
Prove by contradiction. Suppose $L=lim_{x to a} f(x) <0$. Let $0<epsilon <-L$. Then, for $|x-a|$ sufficiently small we get $|f(x)-L| <epsilon$ which implies $f(x) <L+epsilon <0$, a contradiction.
answered Feb 3 at 0:48
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
$endgroup$
add a comment |
$begingroup$
Prove by contradiction. Suppose $L=lim_{x to a} f(x) <0$. Let $0<epsilon <-L$. Then, for $|x-a|$ sufficiently small we get $|f(x)-L| <epsilon$ which implies $f(x) <L+epsilon <0$, a contradiction.
answered Feb 3 at 0:48
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
$endgroup$
add a comment |
$begingroup$
Prove by contradiction. Suppose $L=lim_{x to a} f(x) <0$. Let $0<epsilon <-L$. Then, for $|x-a|$ sufficiently small we get $|f(x)-L| <epsilon$ which implies $f(x) <L+epsilon <0$, a contradiction.
answered Feb 3 at 0:48
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
$endgroup$
Prove by contradiction. Suppose $L=lim_{x to a} f(x) <0$. Let $0<epsilon <-L$. Then, for $|x-a|$ sufficiently small we get $|f(x)-L| <epsilon$ which implies $f(x) <L+epsilon <0$, a contradiction.
answered Feb 3 at 0:48
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
answered Feb 3 at 0:48
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
answered Feb 3 at 0:48
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
answered Feb 3 at 0:48
Kavi Rama MurthyKavi Rama Murthy
74.8k53270
74.8k53270
add a comment |
add a comment |
$begingroup$
Since $f(x)>0$, $-epsilon<f(x)-epsilon$. You have shown that for every $epsilon>0$, there exists $x$ such that $-epsilon<f(x)-epsilonleq l$ implies $lgeq 0$.
answered Feb 3 at 0:16
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
$begingroup$
but since $epsilon>0$, then $-epsilon<0$, so $𝑓(𝑥)−𝜖$ can be negative too, no?
$endgroup$
– dxdydz
Feb 3 at 0:24
$begingroup$
A number which is superior to every negative number is positive, if not, $l<0, l<{lover 2}<0$.
$endgroup$
– Tsemo Aristide
Feb 3 at 0:28
add a comment |
$begingroup$
Since $f(x)>0$, $-epsilon<f(x)-epsilon$. You have shown that for every $epsilon>0$, there exists $x$ such that $-epsilon<f(x)-epsilonleq l$ implies $lgeq 0$.
answered Feb 3 at 0:16
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
$begingroup$
but since $epsilon>0$, then $-epsilon<0$, so $𝑓(𝑥)−𝜖$ can be negative too, no?
$endgroup$
– dxdydz
Feb 3 at 0:24
$begingroup$
A number which is superior to every negative number is positive, if not, $l<0, l<{lover 2}<0$.
$endgroup$
– Tsemo Aristide
Feb 3 at 0:28
add a comment |
$begingroup$
Since $f(x)>0$, $-epsilon<f(x)-epsilon$. You have shown that for every $epsilon>0$, there exists $x$ such that $-epsilon<f(x)-epsilonleq l$ implies $lgeq 0$.
answered Feb 3 at 0:16
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
Since $f(x)>0$, $-epsilon<f(x)-epsilon$. You have shown that for every $epsilon>0$, there exists $x$ such that $-epsilon<f(x)-epsilonleq l$ implies $lgeq 0$.
answered Feb 3 at 0:16
Tsemo AristideTsemo Aristide
60.7k11446
answered Feb 3 at 0:16
Tsemo AristideTsemo Aristide
60.7k11446
answered Feb 3 at 0:16
Tsemo AristideTsemo Aristide
60.7k11446
answered Feb 3 at 0:16
Tsemo AristideTsemo Aristide
60.7k11446
60.7k11446
$begingroup$
but since $epsilon>0$, then $-epsilon<0$, so $𝑓(𝑥)−𝜖$ can be negative too, no?
$endgroup$
– dxdydz
Feb 3 at 0:24
$begingroup$
A number which is superior to every negative number is positive, if not, $l<0, l<{lover 2}<0$.
$endgroup$
– Tsemo Aristide
Feb 3 at 0:28
add a comment |
$begingroup$
but since $epsilon>0$, then $-epsilon<0$, so $𝑓(𝑥)−𝜖$ can be negative too, no?
$endgroup$
– dxdydz
Feb 3 at 0:24
$begingroup$
A number which is superior to every negative number is positive, if not, $l<0, l<{lover 2}<0$.
$endgroup$
– Tsemo Aristide
Feb 3 at 0:28
$begingroup$
but since $epsilon>0$, then $-epsilon<0$, so $𝑓(𝑥)−𝜖$ can be negative too, no?
$endgroup$
– dxdydz
Feb 3 at 0:24
$begingroup$
but since $epsilon>0$, then $-epsilon<0$, so $𝑓(𝑥)−𝜖$ can be negative too, no?
$endgroup$
– dxdydz
Feb 3 at 0:24
$begingroup$
A number which is superior to every negative number is positive, if not, $l<0, l<{lover 2}<0$.
$endgroup$
– Tsemo Aristide
Feb 3 at 0:28
$begingroup$
A number which is superior to every negative number is positive, if not, $l<0, l<{lover 2}<0$.
$endgroup$
– Tsemo Aristide
Feb 3 at 0:28
add a comment |
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