Epsilon-delta proof verification (only finding good delta)
$varepsilon-delta$ proof of $$ lim_{x to -frac{1}{2}} frac{1}{x+1} $$
Here's how I found $delta$:
We got to somehow bound $delta$ with $epsilon$, so let's find $|x + frac{1}{2}|$ in $|frac{1}{x+1} - 2|$. We have:
$$ left |frac{1}{x+1} - 2 right | = left |frac{1}{x+1} - frac{2(x+1)}{x+1} right | = left |frac{1 - 2x - 2}{x+1} right | = left |frac{-2(x + frac{1}{2})}{x+1} right | = frac{2|x + frac{1}{2}|}{|x+1|} $$
So let's see what's happening when $|x + frac{1}{2}| < 1$. We have $ -frac{1}{2} < x < frac{3}{2} $, so upper bound for it is $frac{3}{2}$. We can bound $|x + 1|$ using this information - it wouldn't be smaller than $frac{1}{2}$. Thus $$ frac{2|x + frac{1}{2}|}{|x+1|} < frac{2}{|x+1|} < frac{2}{frac{1}{2}} = 4 $$
Therefore our delta is $delta = min { 1, frac{varepsilon}{4} }$
Is this a good delta?
limits proof-verification epsilon-delta
asked Nov 26 at 17:56
chandx
628
add a comment |
$varepsilon-delta$ proof of $$ lim_{x to -frac{1}{2}} frac{1}{x+1} $$
Here's how I found $delta$:
We got to somehow bound $delta$ with $epsilon$, so let's find $|x + frac{1}{2}|$ in $|frac{1}{x+1} - 2|$. We have:
$$ left |frac{1}{x+1} - 2 right | = left |frac{1}{x+1} - frac{2(x+1)}{x+1} right | = left |frac{1 - 2x - 2}{x+1} right | = left |frac{-2(x + frac{1}{2})}{x+1} right | = frac{2|x + frac{1}{2}|}{|x+1|} $$
So let's see what's happening when $|x + frac{1}{2}| < 1$. We have $ -frac{1}{2} < x < frac{3}{2} $, so upper bound for it is $frac{3}{2}$. We can bound $|x + 1|$ using this information - it wouldn't be smaller than $frac{1}{2}$. Thus $$ frac{2|x + frac{1}{2}|}{|x+1|} < frac{2}{|x+1|} < frac{2}{frac{1}{2}} = 4 $$
Therefore our delta is $delta = min { 1, frac{varepsilon}{4} }$
Is this a good delta?
limits proof-verification epsilon-delta
asked Nov 26 at 17:56
chandx
628
Yes yes and yes
– hamam_Abdallah
Nov 26 at 18:22
add a comment |
$varepsilon-delta$ proof of $$ lim_{x to -frac{1}{2}} frac{1}{x+1} $$
Here's how I found $delta$:
We got to somehow bound $delta$ with $epsilon$, so let's find $|x + frac{1}{2}|$ in $|frac{1}{x+1} - 2|$. We have:
$$ left |frac{1}{x+1} - 2 right | = left |frac{1}{x+1} - frac{2(x+1)}{x+1} right | = left |frac{1 - 2x - 2}{x+1} right | = left |frac{-2(x + frac{1}{2})}{x+1} right | = frac{2|x + frac{1}{2}|}{|x+1|} $$
So let's see what's happening when $|x + frac{1}{2}| < 1$. We have $ -frac{1}{2} < x < frac{3}{2} $, so upper bound for it is $frac{3}{2}$. We can bound $|x + 1|$ using this information - it wouldn't be smaller than $frac{1}{2}$. Thus $$ frac{2|x + frac{1}{2}|}{|x+1|} < frac{2}{|x+1|} < frac{2}{frac{1}{2}} = 4 $$
Therefore our delta is $delta = min { 1, frac{varepsilon}{4} }$
Is this a good delta?
limits proof-verification epsilon-delta
asked Nov 26 at 17:56
chandx
628
$varepsilon-delta$ proof of $$ lim_{x to -frac{1}{2}} frac{1}{x+1} $$
Here's how I found $delta$:
We got to somehow bound $delta$ with $epsilon$, so let's find $|x + frac{1}{2}|$ in $|frac{1}{x+1} - 2|$. We have:
$$ left |frac{1}{x+1} - 2 right | = left |frac{1}{x+1} - frac{2(x+1)}{x+1} right | = left |frac{1 - 2x - 2}{x+1} right | = left |frac{-2(x + frac{1}{2})}{x+1} right | = frac{2|x + frac{1}{2}|}{|x+1|} $$
So let's see what's happening when $|x + frac{1}{2}| < 1$. We have $ -frac{1}{2} < x < frac{3}{2} $, so upper bound for it is $frac{3}{2}$. We can bound $|x + 1|$ using this information - it wouldn't be smaller than $frac{1}{2}$. Thus $$ frac{2|x + frac{1}{2}|}{|x+1|} < frac{2}{|x+1|} < frac{2}{frac{1}{2}} = 4 $$
Therefore our delta is $delta = min { 1, frac{varepsilon}{4} }$
Is this a good delta?
limits proof-verification epsilon-delta
limits proof-verification epsilon-delta
asked Nov 26 at 17:56
chandx
628
asked Nov 26 at 17:56
chandx
628
asked Nov 26 at 17:56
chandx
628
asked Nov 26 at 17:56
chandx
628
asked Nov 26 at 17:56
chandx
628
628
Yes yes and yes
– hamam_Abdallah
Nov 26 at 18:22
add a comment |
Yes yes and yes
– hamam_Abdallah
Nov 26 at 18:22
Yes yes and yes
– hamam_Abdallah
Nov 26 at 18:22
Yes yes and yes
– hamam_Abdallah
Nov 26 at 18:22
add a comment |
2 Answers
2
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oldest
votes
If you pick$|x+0.5|<1$, we have $$-1 < x+0.5< 1$$
$$-1.5<x<0.5$$
then $$-0.5 < x+1<1.5$$
then $|x+1|$ might take value $0$.
Let's make the $delta$ smaller.
We want $|x+0.5|< delta$, $$-delta < x+0.5 < delta$$
$$-0.5-delta < x < -0.5+delta$$
$$0.5-delta < x+1<0.5+delta$$
Let $delta < 0.25$, then $|x+1|>frac14$.
$$frac{2|x+0.5|}{|x+1|}le 8delta$$
answered Nov 26 at 18:05
Siong Thye Goh
98.9k1464116
Oh, those minuses, indeed it may takie value 0. Why you've picked delta < 0.25?
– chandx
Nov 26 at 19:37
so that they don't cross $0$ to better control the denominator.
– Siong Thye Goh
Nov 27 at 0:40
add a comment |
Hint :
By the Triagnle Inequality, one can yield :
$$bigg|x + frac{1}{2}bigg| = Bigg|(x + 1) + bigg(-frac{1}{2}bigg)Bigg|leq |x+1| + 1/2$$
Thus :
$$Bigg|frac{1}{x+1}-2Bigg| leq frac{2|x+1|+1}{|x+1|}=2+frac{1}{|x+1|}$$
So, what can be concluded ?
answered Nov 26 at 18:06
Rebellos
14.3k31245
Well, I think I can't see where is it going, |f(x) - 2| is 2 + something, it can't be arbitrarily small
– chandx
Nov 26 at 19:45
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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votes
If you pick$|x+0.5|<1$, we have $$-1 < x+0.5< 1$$
$$-1.5<x<0.5$$
then $$-0.5 < x+1<1.5$$
then $|x+1|$ might take value $0$.
Let's make the $delta$ smaller.
We want $|x+0.5|< delta$, $$-delta < x+0.5 < delta$$
$$-0.5-delta < x < -0.5+delta$$
$$0.5-delta < x+1<0.5+delta$$
Let $delta < 0.25$, then $|x+1|>frac14$.
$$frac{2|x+0.5|}{|x+1|}le 8delta$$
answered Nov 26 at 18:05
Siong Thye Goh
98.9k1464116
Oh, those minuses, indeed it may takie value 0. Why you've picked delta < 0.25?
– chandx
Nov 26 at 19:37
so that they don't cross $0$ to better control the denominator.
– Siong Thye Goh
Nov 27 at 0:40
add a comment |
If you pick$|x+0.5|<1$, we have $$-1 < x+0.5< 1$$
$$-1.5<x<0.5$$
then $$-0.5 < x+1<1.5$$
then $|x+1|$ might take value $0$.
Let's make the $delta$ smaller.
We want $|x+0.5|< delta$, $$-delta < x+0.5 < delta$$
$$-0.5-delta < x < -0.5+delta$$
$$0.5-delta < x+1<0.5+delta$$
Let $delta < 0.25$, then $|x+1|>frac14$.
$$frac{2|x+0.5|}{|x+1|}le 8delta$$
answered Nov 26 at 18:05
Siong Thye Goh
98.9k1464116
Oh, those minuses, indeed it may takie value 0. Why you've picked delta < 0.25?
– chandx
Nov 26 at 19:37
so that they don't cross $0$ to better control the denominator.
– Siong Thye Goh
Nov 27 at 0:40
add a comment |
If you pick$|x+0.5|<1$, we have $$-1 < x+0.5< 1$$
$$-1.5<x<0.5$$
then $$-0.5 < x+1<1.5$$
then $|x+1|$ might take value $0$.
Let's make the $delta$ smaller.
We want $|x+0.5|< delta$, $$-delta < x+0.5 < delta$$
$$-0.5-delta < x < -0.5+delta$$
$$0.5-delta < x+1<0.5+delta$$
Let $delta < 0.25$, then $|x+1|>frac14$.
$$frac{2|x+0.5|}{|x+1|}le 8delta$$
answered Nov 26 at 18:05
Siong Thye Goh
98.9k1464116
If you pick$|x+0.5|<1$, we have $$-1 < x+0.5< 1$$
$$-1.5<x<0.5$$
then $$-0.5 < x+1<1.5$$
then $|x+1|$ might take value $0$.
Let's make the $delta$ smaller.
We want $|x+0.5|< delta$, $$-delta < x+0.5 < delta$$
$$-0.5-delta < x < -0.5+delta$$
$$0.5-delta < x+1<0.5+delta$$
Let $delta < 0.25$, then $|x+1|>frac14$.
$$frac{2|x+0.5|}{|x+1|}le 8delta$$
answered Nov 26 at 18:05
Siong Thye Goh
98.9k1464116
answered Nov 26 at 18:05
Siong Thye Goh
98.9k1464116
answered Nov 26 at 18:05
Siong Thye Goh
98.9k1464116
answered Nov 26 at 18:05
Siong Thye Goh
98.9k1464116
98.9k1464116
Oh, those minuses, indeed it may takie value 0. Why you've picked delta < 0.25?
– chandx
Nov 26 at 19:37
so that they don't cross $0$ to better control the denominator.
– Siong Thye Goh
Nov 27 at 0:40
add a comment |
Oh, those minuses, indeed it may takie value 0. Why you've picked delta < 0.25?
– chandx
Nov 26 at 19:37
so that they don't cross $0$ to better control the denominator.
– Siong Thye Goh
Nov 27 at 0:40
Oh, those minuses, indeed it may takie value 0. Why you've picked delta < 0.25?
– chandx
Nov 26 at 19:37
Oh, those minuses, indeed it may takie value 0. Why you've picked delta < 0.25?
– chandx
Nov 26 at 19:37
so that they don't cross $0$ to better control the denominator.
– Siong Thye Goh
Nov 27 at 0:40
so that they don't cross $0$ to better control the denominator.
– Siong Thye Goh
Nov 27 at 0:40
add a comment |
Hint :
By the Triagnle Inequality, one can yield :
$$bigg|x + frac{1}{2}bigg| = Bigg|(x + 1) + bigg(-frac{1}{2}bigg)Bigg|leq |x+1| + 1/2$$
Thus :
$$Bigg|frac{1}{x+1}-2Bigg| leq frac{2|x+1|+1}{|x+1|}=2+frac{1}{|x+1|}$$
So, what can be concluded ?
answered Nov 26 at 18:06
Rebellos
14.3k31245
Well, I think I can't see where is it going, |f(x) - 2| is 2 + something, it can't be arbitrarily small
– chandx
Nov 26 at 19:45
add a comment |
Hint :
By the Triagnle Inequality, one can yield :
$$bigg|x + frac{1}{2}bigg| = Bigg|(x + 1) + bigg(-frac{1}{2}bigg)Bigg|leq |x+1| + 1/2$$
Thus :
$$Bigg|frac{1}{x+1}-2Bigg| leq frac{2|x+1|+1}{|x+1|}=2+frac{1}{|x+1|}$$
So, what can be concluded ?
answered Nov 26 at 18:06
Rebellos
14.3k31245
Well, I think I can't see where is it going, |f(x) - 2| is 2 + something, it can't be arbitrarily small
– chandx
Nov 26 at 19:45
add a comment |
Hint :
By the Triagnle Inequality, one can yield :
$$bigg|x + frac{1}{2}bigg| = Bigg|(x + 1) + bigg(-frac{1}{2}bigg)Bigg|leq |x+1| + 1/2$$
Thus :
$$Bigg|frac{1}{x+1}-2Bigg| leq frac{2|x+1|+1}{|x+1|}=2+frac{1}{|x+1|}$$
So, what can be concluded ?
answered Nov 26 at 18:06
Rebellos
14.3k31245
Hint :
By the Triagnle Inequality, one can yield :
$$bigg|x + frac{1}{2}bigg| = Bigg|(x + 1) + bigg(-frac{1}{2}bigg)Bigg|leq |x+1| + 1/2$$
Thus :
$$Bigg|frac{1}{x+1}-2Bigg| leq frac{2|x+1|+1}{|x+1|}=2+frac{1}{|x+1|}$$
So, what can be concluded ?
answered Nov 26 at 18:06
Rebellos
14.3k31245
answered Nov 26 at 18:06
Rebellos
14.3k31245
answered Nov 26 at 18:06
Rebellos
14.3k31245
answered Nov 26 at 18:06
Rebellos
14.3k31245
14.3k31245
Well, I think I can't see where is it going, |f(x) - 2| is 2 + something, it can't be arbitrarily small
– chandx
Nov 26 at 19:45
add a comment |
Well, I think I can't see where is it going, |f(x) - 2| is 2 + something, it can't be arbitrarily small
– chandx
Nov 26 at 19:45
Well, I think I can't see where is it going, |f(x) - 2| is 2 + something, it can't be arbitrarily small
– chandx
Nov 26 at 19:45
Well, I think I can't see where is it going, |f(x) - 2| is 2 + something, it can't be arbitrarily small
– chandx
Nov 26 at 19:45
add a comment |
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Yes yes and yes
– hamam_Abdallah
Nov 26 at 18:22