Mixing
If $f(x) >0$ for all $x$, then $lim_{xto a} f(x) ge 0$
$begingroup$
I met a problem like this:
Suppose $f(x) > 0$ for all $x$ , and also that $lim_{xto a} f(x)$ exists.
a) Show that $lim_{xto a} f(x) ge 0$.
b) Give an example where $lim_{xto a} f(x) = 0$.
I am not sure how to solve it.. in fact I have no idea.
Could anyone give me a hint？
Thanks!!!
real-analysis limits
asked Oct 18 '15 at 5:51
Kristian von RiechstoffenKristian von Riechstoffen
142
$endgroup$
|
show 3 more comments
$begingroup$
I met a problem like this:
Suppose $f(x) > 0$ for all $x$ , and also that $lim_{xto a} f(x)$ exists.
a) Show that $lim_{xto a} f(x) ge 0$.
b) Give an example where $lim_{xto a} f(x) = 0$.
I am not sure how to solve it.. in fact I have no idea.
Could anyone give me a hint？
Thanks!!!
real-analysis limits
asked Oct 18 '15 at 5:51
Kristian von RiechstoffenKristian von Riechstoffen
142
$endgroup$
$begingroup$
Do you mean the original function to be $f(x)ge 0$?
$endgroup$
– Adam Hrankowski
Oct 18 '15 at 5:53
$begingroup$
@AdamHrankowski It wouldn't affect the conclusion.
$endgroup$
– BrianO
Oct 18 '15 at 5:55
$begingroup$
Suppose the limit is $ L <0$. If I give you $epsilon = |L|/2$, can you give me $delta$?
$endgroup$
– littleO
Oct 18 '15 at 5:55
$begingroup$
The question is about the original function, f(x) is always greater than 0 for all x. also, lim x-a f(x) exists, say lim x→a f(x) = L. We want to prove that L is ≥ 0
$endgroup$
– Kristian von Riechstoffen
Oct 18 '15 at 5:56
2
$begingroup$
$ge$ would make b) trivial, so it probably does mean $>$.
$endgroup$
– Ben Millwood
Oct 18 '15 at 5:56
|
show 3 more comments
$begingroup$
I met a problem like this:
Suppose $f(x) > 0$ for all $x$ , and also that $lim_{xto a} f(x)$ exists.
a) Show that $lim_{xto a} f(x) ge 0$.
b) Give an example where $lim_{xto a} f(x) = 0$.
I am not sure how to solve it.. in fact I have no idea.
Could anyone give me a hint？
Thanks!!!
real-analysis limits
asked Oct 18 '15 at 5:51
Kristian von RiechstoffenKristian von Riechstoffen
142
$endgroup$
I met a problem like this:
Suppose $f(x) > 0$ for all $x$ , and also that $lim_{xto a} f(x)$ exists.
a) Show that $lim_{xto a} f(x) ge 0$.
b) Give an example where $lim_{xto a} f(x) = 0$.
I am not sure how to solve it.. in fact I have no idea.
Could anyone give me a hint？
Thanks!!!
real-analysis limits
real-analysis limits
asked Oct 18 '15 at 5:51
Kristian von RiechstoffenKristian von Riechstoffen
142
asked Oct 18 '15 at 5:51
Kristian von RiechstoffenKristian von Riechstoffen
142
edited Oct 18 '15 at 5:57
user99914
asked Oct 18 '15 at 5:51
Kristian von RiechstoffenKristian von Riechstoffen
142
asked Oct 18 '15 at 5:51
Kristian von RiechstoffenKristian von Riechstoffen
142
asked Oct 18 '15 at 5:51
Kristian von RiechstoffenKristian von Riechstoffen
142
142
$begingroup$
Do you mean the original function to be $f(x)ge 0$?
$endgroup$
– Adam Hrankowski
Oct 18 '15 at 5:53
$begingroup$
@AdamHrankowski It wouldn't affect the conclusion.
$endgroup$
– BrianO
Oct 18 '15 at 5:55
$begingroup$
Suppose the limit is $ L <0$. If I give you $epsilon = |L|/2$, can you give me $delta$?
$endgroup$
– littleO
Oct 18 '15 at 5:55
$begingroup$
The question is about the original function, f(x) is always greater than 0 for all x. also, lim x-a f(x) exists, say lim x→a f(x) = L. We want to prove that L is ≥ 0
$endgroup$
– Kristian von Riechstoffen
Oct 18 '15 at 5:56
2
$begingroup$
$ge$ would make b) trivial, so it probably does mean $>$.
$endgroup$
– Ben Millwood
Oct 18 '15 at 5:56
|
show 3 more comments
$begingroup$
Do you mean the original function to be $f(x)ge 0$?
$endgroup$
– Adam Hrankowski
Oct 18 '15 at 5:53
$begingroup$
@AdamHrankowski It wouldn't affect the conclusion.
$endgroup$
– BrianO
Oct 18 '15 at 5:55
$begingroup$
Suppose the limit is $ L <0$. If I give you $epsilon = |L|/2$, can you give me $delta$?
$endgroup$
– littleO
Oct 18 '15 at 5:55
$begingroup$
The question is about the original function, f(x) is always greater than 0 for all x. also, lim x-a f(x) exists, say lim x→a f(x) = L. We want to prove that L is ≥ 0
$endgroup$
– Kristian von Riechstoffen
Oct 18 '15 at 5:56
2
$begingroup$
$ge$ would make b) trivial, so it probably does mean $>$.
$endgroup$
– Ben Millwood
Oct 18 '15 at 5:56
$begingroup$
Do you mean the original function to be $f(x)ge 0$?
$endgroup$
– Adam Hrankowski
Oct 18 '15 at 5:53
$begingroup$
Do you mean the original function to be $f(x)ge 0$?
$endgroup$
– Adam Hrankowski
Oct 18 '15 at 5:53
$begingroup$
@AdamHrankowski It wouldn't affect the conclusion.
$endgroup$
– BrianO
Oct 18 '15 at 5:55
$begingroup$
@AdamHrankowski It wouldn't affect the conclusion.
$endgroup$
– BrianO
Oct 18 '15 at 5:55
$begingroup$
Suppose the limit is $ L <0$. If I give you $epsilon = |L|/2$, can you give me $delta$?
$endgroup$
– littleO
Oct 18 '15 at 5:55
$begingroup$
Suppose the limit is $ L <0$. If I give you $epsilon = |L|/2$, can you give me $delta$?
$endgroup$
– littleO
Oct 18 '15 at 5:55
$begingroup$
The question is about the original function, f(x) is always greater than 0 for all x. also, lim x-a f(x) exists, say lim x→a f(x) = L. We want to prove that L is ≥ 0
$endgroup$
– Kristian von Riechstoffen
Oct 18 '15 at 5:56
$begingroup$
The question is about the original function, f(x) is always greater than 0 for all x. also, lim x-a f(x) exists, say lim x→a f(x) = L. We want to prove that L is ≥ 0
$endgroup$
– Kristian von Riechstoffen
Oct 18 '15 at 5:56
2
2
$begingroup$
$ge$ would make b) trivial, so it probably does mean $>$.
$endgroup$
– Ben Millwood
Oct 18 '15 at 5:56
$begingroup$
$ge$ would make b) trivial, so it probably does mean $>$.
$endgroup$
– Ben Millwood
Oct 18 '15 at 5:56
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Let $L$ be the limit, you wish to show: $L geq 0$. You could prove this directly, but most "students" would choose the indirect proof. So assume $L < 0$.Thus choose $epsilon = dfrac{-L}{4} > 0$, then there is a $delta > 0$ such that if $0 < |x-a| < delta$, then $|f(x) - L| < dfrac{-L}{4} Rightarrow f(x) - L leq |f(x)-L| < dfrac{-L}{4} Rightarrow f(x) < L-dfrac{L}{4}= dfrac{3L}{4} < 0$, contradiction. Thus $L geq 0$. An example of such a function is $f(x) = e^{-dfrac{1}{x^2}}, x neq 0$,and $1, x = 0$ has $displaystyle lim_{x to 0} f(x)= 0$
answered Oct 18 '15 at 6:43
DeepSeaDeepSea
71.5k54488
$endgroup$
$begingroup$
Thank you so much, ! think that's a great answer!
$endgroup$
– Kristian von Riechstoffen
Oct 18 '15 at 17:06
add a comment |
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1 Answer
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$begingroup$
Let $L$ be the limit, you wish to show: $L geq 0$. You could prove this directly, but most "students" would choose the indirect proof. So assume $L < 0$.Thus choose $epsilon = dfrac{-L}{4} > 0$, then there is a $delta > 0$ such that if $0 < |x-a| < delta$, then $|f(x) - L| < dfrac{-L}{4} Rightarrow f(x) - L leq |f(x)-L| < dfrac{-L}{4} Rightarrow f(x) < L-dfrac{L}{4}= dfrac{3L}{4} < 0$, contradiction. Thus $L geq 0$. An example of such a function is $f(x) = e^{-dfrac{1}{x^2}}, x neq 0$,and $1, x = 0$ has $displaystyle lim_{x to 0} f(x)= 0$
answered Oct 18 '15 at 6:43
DeepSeaDeepSea
71.5k54488
$endgroup$
$begingroup$
Thank you so much, ! think that's a great answer!
$endgroup$
– Kristian von Riechstoffen
Oct 18 '15 at 17:06
add a comment |
$begingroup$
Let $L$ be the limit, you wish to show: $L geq 0$. You could prove this directly, but most "students" would choose the indirect proof. So assume $L < 0$.Thus choose $epsilon = dfrac{-L}{4} > 0$, then there is a $delta > 0$ such that if $0 < |x-a| < delta$, then $|f(x) - L| < dfrac{-L}{4} Rightarrow f(x) - L leq |f(x)-L| < dfrac{-L}{4} Rightarrow f(x) < L-dfrac{L}{4}= dfrac{3L}{4} < 0$, contradiction. Thus $L geq 0$. An example of such a function is $f(x) = e^{-dfrac{1}{x^2}}, x neq 0$,and $1, x = 0$ has $displaystyle lim_{x to 0} f(x)= 0$
answered Oct 18 '15 at 6:43
DeepSeaDeepSea
71.5k54488
$endgroup$
$begingroup$
Thank you so much, ! think that's a great answer!
$endgroup$
– Kristian von Riechstoffen
Oct 18 '15 at 17:06
add a comment |
$begingroup$
Let $L$ be the limit, you wish to show: $L geq 0$. You could prove this directly, but most "students" would choose the indirect proof. So assume $L < 0$.Thus choose $epsilon = dfrac{-L}{4} > 0$, then there is a $delta > 0$ such that if $0 < |x-a| < delta$, then $|f(x) - L| < dfrac{-L}{4} Rightarrow f(x) - L leq |f(x)-L| < dfrac{-L}{4} Rightarrow f(x) < L-dfrac{L}{4}= dfrac{3L}{4} < 0$, contradiction. Thus $L geq 0$. An example of such a function is $f(x) = e^{-dfrac{1}{x^2}}, x neq 0$,and $1, x = 0$ has $displaystyle lim_{x to 0} f(x)= 0$
answered Oct 18 '15 at 6:43
DeepSeaDeepSea
71.5k54488
$endgroup$
Let $L$ be the limit, you wish to show: $L geq 0$. You could prove this directly, but most "students" would choose the indirect proof. So assume $L < 0$.Thus choose $epsilon = dfrac{-L}{4} > 0$, then there is a $delta > 0$ such that if $0 < |x-a| < delta$, then $|f(x) - L| < dfrac{-L}{4} Rightarrow f(x) - L leq |f(x)-L| < dfrac{-L}{4} Rightarrow f(x) < L-dfrac{L}{4}= dfrac{3L}{4} < 0$, contradiction. Thus $L geq 0$. An example of such a function is $f(x) = e^{-dfrac{1}{x^2}}, x neq 0$,and $1, x = 0$ has $displaystyle lim_{x to 0} f(x)= 0$
answered Oct 18 '15 at 6:43
DeepSeaDeepSea
71.5k54488
answered Oct 18 '15 at 6:43
DeepSeaDeepSea
71.5k54488
answered Oct 18 '15 at 6:43
DeepSeaDeepSea
71.5k54488
answered Oct 18 '15 at 6:43
DeepSeaDeepSea
71.5k54488
71.5k54488
$begingroup$
Thank you so much, ! think that's a great answer!
$endgroup$
– Kristian von Riechstoffen
Oct 18 '15 at 17:06
add a comment |
$begingroup$
Thank you so much, ! think that's a great answer!
$endgroup$
– Kristian von Riechstoffen
Oct 18 '15 at 17:06
$begingroup$
Thank you so much, ! think that's a great answer!
$endgroup$
– Kristian von Riechstoffen
Oct 18 '15 at 17:06
$begingroup$
Thank you so much, ! think that's a great answer!
$endgroup$
– Kristian von Riechstoffen
Oct 18 '15 at 17:06
add a comment |
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$begingroup$
Do you mean the original function to be $f(x)ge 0$?
$endgroup$
– Adam Hrankowski
Oct 18 '15 at 5:53
$begingroup$
@AdamHrankowski It wouldn't affect the conclusion.
$endgroup$
– BrianO
Oct 18 '15 at 5:55
$begingroup$
Suppose the limit is $ L <0$. If I give you $epsilon = |L|/2$, can you give me $delta$?
$endgroup$
– littleO
Oct 18 '15 at 5:55
$begingroup$
The question is about the original function, f(x) is always greater than 0 for all x. also, lim x-a f(x) exists, say lim x→a f(x) = L. We want to prove that L is ≥ 0
$endgroup$
– Kristian von Riechstoffen
Oct 18 '15 at 5:56
2
$begingroup$
$ge$ would make b) trivial, so it probably does mean $>$.
$endgroup$
– Ben Millwood
Oct 18 '15 at 5:56