Prove that $lim_{x to a} f(x) = infty$ iff $lim_{x to a} frac{1}{f(x)} = 0$
Let $a in mathbb{R}$ and prove that, using the $epsilon-delta$ definition of the limit, $displaystyle lim_{x to a} f(x) = infty$ iff $displaystyle lim_{x to a} frac{1}{f(x)} = 0$ and that there is $delta > 0$ such that $f(x) > 0$ whenever $0 < |x-a| < delta$.
For the first question, we are given that $$forall M>0, exists delta quad 0<|x-a|<delta implies quad f(x) > M$$ and need to show this is equivalent to $$forall epsilon, exists delta quad 0<|x-a|<delta implies quad left |dfrac{1}{f(x)} right |< epsilon.$$ I am not to sure how to show this part. As for the second part of the question, couldn't we just have $f(x) = -5$ and that never be true?
calculus limits
edited Nov 27 at 12:14
Martin Sleziak
44.7k7115270
asked Mar 7 '16 at 1:10
Puzzled417
3,085527
add a comment |
Let $a in mathbb{R}$ and prove that, using the $epsilon-delta$ definition of the limit, $displaystyle lim_{x to a} f(x) = infty$ iff $displaystyle lim_{x to a} frac{1}{f(x)} = 0$ and that there is $delta > 0$ such that $f(x) > 0$ whenever $0 < |x-a| < delta$.
For the first question, we are given that $$forall M>0, exists delta quad 0<|x-a|<delta implies quad f(x) > M$$ and need to show this is equivalent to $$forall epsilon, exists delta quad 0<|x-a|<delta implies quad left |dfrac{1}{f(x)} right |< epsilon.$$ I am not to sure how to show this part. As for the second part of the question, couldn't we just have $f(x) = -5$ and that never be true?
calculus limits
edited Nov 27 at 12:14
Martin Sleziak
44.7k7115270
asked Mar 7 '16 at 1:10
Puzzled417
3,085527
Actually that's not right @ThomasAndrews. I think your example is wrong this should work for any function since it approaches infinity as $a$ goes to infinity.
– Puzzled417
Mar 7 '16 at 2:11
Remember we are talking about $f(x)$ approaching infinity as $a$ goes to infinity, which it does in your example. But the $infty$ I define here means $+infty$ so this works for any function.
– Puzzled417
Mar 7 '16 at 2:16
add a comment |
Let $a in mathbb{R}$ and prove that, using the $epsilon-delta$ definition of the limit, $displaystyle lim_{x to a} f(x) = infty$ iff $displaystyle lim_{x to a} frac{1}{f(x)} = 0$ and that there is $delta > 0$ such that $f(x) > 0$ whenever $0 < |x-a| < delta$.
For the first question, we are given that $$forall M>0, exists delta quad 0<|x-a|<delta implies quad f(x) > M$$ and need to show this is equivalent to $$forall epsilon, exists delta quad 0<|x-a|<delta implies quad left |dfrac{1}{f(x)} right |< epsilon.$$ I am not to sure how to show this part. As for the second part of the question, couldn't we just have $f(x) = -5$ and that never be true?
calculus limits
edited Nov 27 at 12:14
Martin Sleziak
44.7k7115270
asked Mar 7 '16 at 1:10
Puzzled417
3,085527
Let $a in mathbb{R}$ and prove that, using the $epsilon-delta$ definition of the limit, $displaystyle lim_{x to a} f(x) = infty$ iff $displaystyle lim_{x to a} frac{1}{f(x)} = 0$ and that there is $delta > 0$ such that $f(x) > 0$ whenever $0 < |x-a| < delta$.
For the first question, we are given that $$forall M>0, exists delta quad 0<|x-a|<delta implies quad f(x) > M$$ and need to show this is equivalent to $$forall epsilon, exists delta quad 0<|x-a|<delta implies quad left |dfrac{1}{f(x)} right |< epsilon.$$ I am not to sure how to show this part. As for the second part of the question, couldn't we just have $f(x) = -5$ and that never be true?
calculus limits
calculus limits
edited Nov 27 at 12:14
Martin Sleziak
44.7k7115270
asked Mar 7 '16 at 1:10
Puzzled417
3,085527
edited Nov 27 at 12:14
Martin Sleziak
44.7k7115270
asked Mar 7 '16 at 1:10
Puzzled417
3,085527
edited Nov 27 at 12:14
Martin Sleziak
44.7k7115270
edited Nov 27 at 12:14
Martin Sleziak
44.7k7115270
edited Nov 27 at 12:14
Martin Sleziak
44.7k7115270
44.7k7115270
asked Mar 7 '16 at 1:10
Puzzled417
3,085527
asked Mar 7 '16 at 1:10
Puzzled417
3,085527
asked Mar 7 '16 at 1:10
Puzzled417
3,085527
3,085527
Actually that's not right @ThomasAndrews. I think your example is wrong this should work for any function since it approaches infinity as $a$ goes to infinity.
– Puzzled417
Mar 7 '16 at 2:11
Remember we are talking about $f(x)$ approaching infinity as $a$ goes to infinity, which it does in your example. But the $infty$ I define here means $+infty$ so this works for any function.
– Puzzled417
Mar 7 '16 at 2:16
add a comment |
Actually that's not right @ThomasAndrews. I think your example is wrong this should work for any function since it approaches infinity as $a$ goes to infinity.
– Puzzled417
Mar 7 '16 at 2:11
Remember we are talking about $f(x)$ approaching infinity as $a$ goes to infinity, which it does in your example. But the $infty$ I define here means $+infty$ so this works for any function.
– Puzzled417
Mar 7 '16 at 2:16
Actually that's not right @ThomasAndrews. I think your example is wrong this should work for any function since it approaches infinity as $a$ goes to infinity.
– Puzzled417
Mar 7 '16 at 2:11
Actually that's not right @ThomasAndrews. I think your example is wrong this should work for any function since it approaches infinity as $a$ goes to infinity.
– Puzzled417
Mar 7 '16 at 2:11
Remember we are talking about $f(x)$ approaching infinity as $a$ goes to infinity, which it does in your example. But the $infty$ I define here means $+infty$ so this works for any function.
– Puzzled417
Mar 7 '16 at 2:16
Remember we are talking about $f(x)$ approaching infinity as $a$ goes to infinity, which it does in your example. But the $infty$ I define here means $+infty$ so this works for any function.
– Puzzled417
Mar 7 '16 at 2:16
add a comment |
3 Answers
3
active
oldest
votes
Here is the first part: Let $varepsilon>0$. Then, by assumption, $exists delta>0$ such that $forall x: |x-a|<deltaimplies f(x)>frac{1}{varepsilon}$. Thus, $frac{1}{f(x)}<varepsilon$ and since $varepsilon$ was arbitrary, $limlimits_{xrightarrow a}frac{1}{f(x)}=0$.
answered Mar 7 '16 at 1:16
Tony S.F.
3,2222928
add a comment |
So let some $epsilon > 0$. We need to find a $delta$ such that when $|x-a|<delta$, we would have $left|frac{1}{f(x)}right| < epsilon$.
Let $M = 1/epsilon$. Now there must exist a $delta$ such that when $|x-a|<delta$, we have $f(x) > M = 1/epsilon$.
Can you finish this by inverting the last inequality?
answered Mar 7 '16 at 1:16
gt6989b
33k22452
We have $frac{1}{f(x)} < epsilon$. We can't take the absolute value of both sides, can we?
– Puzzled417
Mar 7 '16 at 1:31
@Puzzled417 But $M = 1/epsilon > 0$ and so $f(x) > 0$ too
– gt6989b
Mar 7 '16 at 1:37
What about the second statement in the question?
– Puzzled417
Mar 7 '16 at 1:41
What about the $f(x) > 0$ part?
– Puzzled417
Mar 7 '16 at 2:31
@Puzzled417 it is talking about the same $f$ in the second part...
– gt6989b
Mar 9 '16 at 17:23
add a comment |
For the direct side of the theorem as you have stated we have$$forall M>0, exists delta quad 0<|x-a|<delta implies quad f(x) > M$$define $M={1over epsilon}$ with $epsilon>0$. Therefore the above logical statement is equivalent to $$forall epsilon>0, exists delta quad 0<|x-a|<delta implies quad f(x) > {1over epsilon}to 0<{1over f(x)}<epsilon$$which is the direct definition of limit to zero. Here we do not need the second condition to hold as a necessary condition while as a necessary condition for the converse we must have that the function must be positive on at least on neighborhood of $a$. The second condition is in fact a formal mathematical expression of this. Sinilarly in the proof of converse, you only need to take $epsilon$ sufficiently small such that $delta<delta_0$ where $delta_0$ satisfies the second condition.
answered Nov 27 at 13:03
Mostafa Ayaz
13.6k3836
add a comment |
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3 Answers
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3 Answers
3
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oldest
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Here is the first part: Let $varepsilon>0$. Then, by assumption, $exists delta>0$ such that $forall x: |x-a|<deltaimplies f(x)>frac{1}{varepsilon}$. Thus, $frac{1}{f(x)}<varepsilon$ and since $varepsilon$ was arbitrary, $limlimits_{xrightarrow a}frac{1}{f(x)}=0$.
answered Mar 7 '16 at 1:16
Tony S.F.
3,2222928
add a comment |
Here is the first part: Let $varepsilon>0$. Then, by assumption, $exists delta>0$ such that $forall x: |x-a|<deltaimplies f(x)>frac{1}{varepsilon}$. Thus, $frac{1}{f(x)}<varepsilon$ and since $varepsilon$ was arbitrary, $limlimits_{xrightarrow a}frac{1}{f(x)}=0$.
answered Mar 7 '16 at 1:16
Tony S.F.
3,2222928
add a comment |
Here is the first part: Let $varepsilon>0$. Then, by assumption, $exists delta>0$ such that $forall x: |x-a|<deltaimplies f(x)>frac{1}{varepsilon}$. Thus, $frac{1}{f(x)}<varepsilon$ and since $varepsilon$ was arbitrary, $limlimits_{xrightarrow a}frac{1}{f(x)}=0$.
answered Mar 7 '16 at 1:16
Tony S.F.
3,2222928
Here is the first part: Let $varepsilon>0$. Then, by assumption, $exists delta>0$ such that $forall x: |x-a|<deltaimplies f(x)>frac{1}{varepsilon}$. Thus, $frac{1}{f(x)}<varepsilon$ and since $varepsilon$ was arbitrary, $limlimits_{xrightarrow a}frac{1}{f(x)}=0$.
answered Mar 7 '16 at 1:16
Tony S.F.
3,2222928
answered Mar 7 '16 at 1:16
Tony S.F.
3,2222928
answered Mar 7 '16 at 1:16
Tony S.F.
3,2222928
answered Mar 7 '16 at 1:16
Tony S.F.
3,2222928
3,2222928
add a comment |
add a comment |
So let some $epsilon > 0$. We need to find a $delta$ such that when $|x-a|<delta$, we would have $left|frac{1}{f(x)}right| < epsilon$.
Let $M = 1/epsilon$. Now there must exist a $delta$ such that when $|x-a|<delta$, we have $f(x) > M = 1/epsilon$.
Can you finish this by inverting the last inequality?
answered Mar 7 '16 at 1:16
gt6989b
33k22452
We have $frac{1}{f(x)} < epsilon$. We can't take the absolute value of both sides, can we?
– Puzzled417
Mar 7 '16 at 1:31
@Puzzled417 But $M = 1/epsilon > 0$ and so $f(x) > 0$ too
– gt6989b
Mar 7 '16 at 1:37
What about the second statement in the question?
– Puzzled417
Mar 7 '16 at 1:41
What about the $f(x) > 0$ part?
– Puzzled417
Mar 7 '16 at 2:31
@Puzzled417 it is talking about the same $f$ in the second part...
– gt6989b
Mar 9 '16 at 17:23
add a comment |
So let some $epsilon > 0$. We need to find a $delta$ such that when $|x-a|<delta$, we would have $left|frac{1}{f(x)}right| < epsilon$.
Let $M = 1/epsilon$. Now there must exist a $delta$ such that when $|x-a|<delta$, we have $f(x) > M = 1/epsilon$.
Can you finish this by inverting the last inequality?
answered Mar 7 '16 at 1:16
gt6989b
33k22452
We have $frac{1}{f(x)} < epsilon$. We can't take the absolute value of both sides, can we?
– Puzzled417
Mar 7 '16 at 1:31
@Puzzled417 But $M = 1/epsilon > 0$ and so $f(x) > 0$ too
– gt6989b
Mar 7 '16 at 1:37
What about the second statement in the question?
– Puzzled417
Mar 7 '16 at 1:41
What about the $f(x) > 0$ part?
– Puzzled417
Mar 7 '16 at 2:31
@Puzzled417 it is talking about the same $f$ in the second part...
– gt6989b
Mar 9 '16 at 17:23
add a comment |
So let some $epsilon > 0$. We need to find a $delta$ such that when $|x-a|<delta$, we would have $left|frac{1}{f(x)}right| < epsilon$.
Let $M = 1/epsilon$. Now there must exist a $delta$ such that when $|x-a|<delta$, we have $f(x) > M = 1/epsilon$.
Can you finish this by inverting the last inequality?
answered Mar 7 '16 at 1:16
gt6989b
33k22452
So let some $epsilon > 0$. We need to find a $delta$ such that when $|x-a|<delta$, we would have $left|frac{1}{f(x)}right| < epsilon$.
Let $M = 1/epsilon$. Now there must exist a $delta$ such that when $|x-a|<delta$, we have $f(x) > M = 1/epsilon$.
Can you finish this by inverting the last inequality?
answered Mar 7 '16 at 1:16
gt6989b
33k22452
answered Mar 7 '16 at 1:16
gt6989b
33k22452
answered Mar 7 '16 at 1:16
gt6989b
33k22452
answered Mar 7 '16 at 1:16
gt6989b
33k22452
33k22452
We have $frac{1}{f(x)} < epsilon$. We can't take the absolute value of both sides, can we?
– Puzzled417
Mar 7 '16 at 1:31
@Puzzled417 But $M = 1/epsilon > 0$ and so $f(x) > 0$ too
– gt6989b
Mar 7 '16 at 1:37
What about the second statement in the question?
– Puzzled417
Mar 7 '16 at 1:41
What about the $f(x) > 0$ part?
– Puzzled417
Mar 7 '16 at 2:31
@Puzzled417 it is talking about the same $f$ in the second part...
– gt6989b
Mar 9 '16 at 17:23
add a comment |
We have $frac{1}{f(x)} < epsilon$. We can't take the absolute value of both sides, can we?
– Puzzled417
Mar 7 '16 at 1:31
@Puzzled417 But $M = 1/epsilon > 0$ and so $f(x) > 0$ too
– gt6989b
Mar 7 '16 at 1:37
What about the second statement in the question?
– Puzzled417
Mar 7 '16 at 1:41
What about the $f(x) > 0$ part?
– Puzzled417
Mar 7 '16 at 2:31
@Puzzled417 it is talking about the same $f$ in the second part...
– gt6989b
Mar 9 '16 at 17:23
We have $frac{1}{f(x)} < epsilon$. We can't take the absolute value of both sides, can we?
– Puzzled417
Mar 7 '16 at 1:31
We have $frac{1}{f(x)} < epsilon$. We can't take the absolute value of both sides, can we?
– Puzzled417
Mar 7 '16 at 1:31
@Puzzled417 But $M = 1/epsilon > 0$ and so $f(x) > 0$ too
– gt6989b
Mar 7 '16 at 1:37
@Puzzled417 But $M = 1/epsilon > 0$ and so $f(x) > 0$ too
– gt6989b
Mar 7 '16 at 1:37
What about the second statement in the question?
– Puzzled417
Mar 7 '16 at 1:41
What about the second statement in the question?
– Puzzled417
Mar 7 '16 at 1:41
What about the $f(x) > 0$ part?
– Puzzled417
Mar 7 '16 at 2:31
What about the $f(x) > 0$ part?
– Puzzled417
Mar 7 '16 at 2:31
@Puzzled417 it is talking about the same $f$ in the second part...
– gt6989b
Mar 9 '16 at 17:23
@Puzzled417 it is talking about the same $f$ in the second part...
– gt6989b
Mar 9 '16 at 17:23
add a comment |
For the direct side of the theorem as you have stated we have$$forall M>0, exists delta quad 0<|x-a|<delta implies quad f(x) > M$$define $M={1over epsilon}$ with $epsilon>0$. Therefore the above logical statement is equivalent to $$forall epsilon>0, exists delta quad 0<|x-a|<delta implies quad f(x) > {1over epsilon}to 0<{1over f(x)}<epsilon$$which is the direct definition of limit to zero. Here we do not need the second condition to hold as a necessary condition while as a necessary condition for the converse we must have that the function must be positive on at least on neighborhood of $a$. The second condition is in fact a formal mathematical expression of this. Sinilarly in the proof of converse, you only need to take $epsilon$ sufficiently small such that $delta<delta_0$ where $delta_0$ satisfies the second condition.
answered Nov 27 at 13:03
Mostafa Ayaz
13.6k3836
add a comment |
For the direct side of the theorem as you have stated we have$$forall M>0, exists delta quad 0<|x-a|<delta implies quad f(x) > M$$define $M={1over epsilon}$ with $epsilon>0$. Therefore the above logical statement is equivalent to $$forall epsilon>0, exists delta quad 0<|x-a|<delta implies quad f(x) > {1over epsilon}to 0<{1over f(x)}<epsilon$$which is the direct definition of limit to zero. Here we do not need the second condition to hold as a necessary condition while as a necessary condition for the converse we must have that the function must be positive on at least on neighborhood of $a$. The second condition is in fact a formal mathematical expression of this. Sinilarly in the proof of converse, you only need to take $epsilon$ sufficiently small such that $delta<delta_0$ where $delta_0$ satisfies the second condition.
answered Nov 27 at 13:03
Mostafa Ayaz
13.6k3836
add a comment |
For the direct side of the theorem as you have stated we have$$forall M>0, exists delta quad 0<|x-a|<delta implies quad f(x) > M$$define $M={1over epsilon}$ with $epsilon>0$. Therefore the above logical statement is equivalent to $$forall epsilon>0, exists delta quad 0<|x-a|<delta implies quad f(x) > {1over epsilon}to 0<{1over f(x)}<epsilon$$which is the direct definition of limit to zero. Here we do not need the second condition to hold as a necessary condition while as a necessary condition for the converse we must have that the function must be positive on at least on neighborhood of $a$. The second condition is in fact a formal mathematical expression of this. Sinilarly in the proof of converse, you only need to take $epsilon$ sufficiently small such that $delta<delta_0$ where $delta_0$ satisfies the second condition.
answered Nov 27 at 13:03
Mostafa Ayaz
13.6k3836
For the direct side of the theorem as you have stated we have$$forall M>0, exists delta quad 0<|x-a|<delta implies quad f(x) > M$$define $M={1over epsilon}$ with $epsilon>0$. Therefore the above logical statement is equivalent to $$forall epsilon>0, exists delta quad 0<|x-a|<delta implies quad f(x) > {1over epsilon}to 0<{1over f(x)}<epsilon$$which is the direct definition of limit to zero. Here we do not need the second condition to hold as a necessary condition while as a necessary condition for the converse we must have that the function must be positive on at least on neighborhood of $a$. The second condition is in fact a formal mathematical expression of this. Sinilarly in the proof of converse, you only need to take $epsilon$ sufficiently small such that $delta<delta_0$ where $delta_0$ satisfies the second condition.
answered Nov 27 at 13:03
Mostafa Ayaz
13.6k3836
answered Nov 27 at 13:03
Mostafa Ayaz
13.6k3836
answered Nov 27 at 13:03
Mostafa Ayaz
13.6k3836
answered Nov 27 at 13:03
Mostafa Ayaz
13.6k3836
13.6k3836
add a comment |
add a comment |
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Actually that's not right @ThomasAndrews. I think your example is wrong this should work for any function since it approaches infinity as $a$ goes to infinity.
– Puzzled417
Mar 7 '16 at 2:11
Remember we are talking about $f(x)$ approaching infinity as $a$ goes to infinity, which it does in your example. But the $infty$ I define here means $+infty$ so this works for any function.
– Puzzled417
Mar 7 '16 at 2:16