Mixing
How can I solve this graph using the precise definition of limits?
$begingroup$
Use the given graph of $f(x)=sqrt{x}$ to find a number $delta$ such that
if $|x-4|<delta$ then $|sqrt{x}-2|<0.4$
This precise definition of a limit has been giving me a lot of trouble, but so far I have the left value of x $f(x)=1.6$ $sqrt{x}=1.6$ $x=2.56$, and the right value $f(x)=2.4$ $sqrt{x}=2.4$ $x=5.76$
Now, I am stuck because I don't know what to do with this information. I think my main problem is that I don't understand this concept in general. I've been going through my textbook and the problems, I still don't get it. So far I've been able to do questions that already have all of the values of $x$ and $f(x)$ on the graph, but only because I have remembered what steps I'm supposed to take, not that I understand what I'm doing (not even the tiniest bit).
What are my next steps?
calculus limits
asked Jun 16 '15 at 17:48
matryoshkamatryoshka
5543720
$endgroup$
add a comment |
$begingroup$
Use the given graph of $f(x)=sqrt{x}$ to find a number $delta$ such that
if $|x-4|<delta$ then $|sqrt{x}-2|<0.4$
This precise definition of a limit has been giving me a lot of trouble, but so far I have the left value of x $f(x)=1.6$ $sqrt{x}=1.6$ $x=2.56$, and the right value $f(x)=2.4$ $sqrt{x}=2.4$ $x=5.76$
Now, I am stuck because I don't know what to do with this information. I think my main problem is that I don't understand this concept in general. I've been going through my textbook and the problems, I still don't get it. So far I've been able to do questions that already have all of the values of $x$ and $f(x)$ on the graph, but only because I have remembered what steps I'm supposed to take, not that I understand what I'm doing (not even the tiniest bit).
What are my next steps?
calculus limits
asked Jun 16 '15 at 17:48
matryoshkamatryoshka
5543720
$endgroup$
add a comment |
$begingroup$
Use the given graph of $f(x)=sqrt{x}$ to find a number $delta$ such that
if $|x-4|<delta$ then $|sqrt{x}-2|<0.4$
This precise definition of a limit has been giving me a lot of trouble, but so far I have the left value of x $f(x)=1.6$ $sqrt{x}=1.6$ $x=2.56$, and the right value $f(x)=2.4$ $sqrt{x}=2.4$ $x=5.76$
Now, I am stuck because I don't know what to do with this information. I think my main problem is that I don't understand this concept in general. I've been going through my textbook and the problems, I still don't get it. So far I've been able to do questions that already have all of the values of $x$ and $f(x)$ on the graph, but only because I have remembered what steps I'm supposed to take, not that I understand what I'm doing (not even the tiniest bit).
What are my next steps?
calculus limits
asked Jun 16 '15 at 17:48
matryoshkamatryoshka
5543720
$endgroup$
Use the given graph of $f(x)=sqrt{x}$ to find a number $delta$ such that
if $|x-4|<delta$ then $|sqrt{x}-2|<0.4$
This precise definition of a limit has been giving me a lot of trouble, but so far I have the left value of x $f(x)=1.6$ $sqrt{x}=1.6$ $x=2.56$, and the right value $f(x)=2.4$ $sqrt{x}=2.4$ $x=5.76$
Now, I am stuck because I don't know what to do with this information. I think my main problem is that I don't understand this concept in general. I've been going through my textbook and the problems, I still don't get it. So far I've been able to do questions that already have all of the values of $x$ and $f(x)$ on the graph, but only because I have remembered what steps I'm supposed to take, not that I understand what I'm doing (not even the tiniest bit).
What are my next steps?
calculus limits
calculus limits
asked Jun 16 '15 at 17:48
matryoshkamatryoshka
5543720
asked Jun 16 '15 at 17:48
matryoshkamatryoshka
5543720
asked Jun 16 '15 at 17:48
matryoshkamatryoshka
5543720
asked Jun 16 '15 at 17:48
matryoshkamatryoshka
5543720
asked Jun 16 '15 at 17:48
matryoshkamatryoshka
5543720
5543720
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Take $delta=4-2.56$, so each element in the interval $]4-delta,4+delta[$ is taken, by $f$, into the $0.4$-neighbourhood of $2$.
answered Jun 16 '15 at 17:54
janmarqzjanmarqz
6,20241630
$endgroup$
$begingroup$
Am I trying to find a number $delta$ that is close to 4, so the limit is brought close to a $f(x)$ value that is less than 0.4? I apologize in advance is this question doesn't make much sense, this concept has been really hard for me to wrap my head around.
$endgroup$
– matryoshka
Jun 16 '15 at 18:11
$begingroup$
you are searching for $delta$ to make a $delta$-neighbourhood of $4$ such that for each point in it, they are carried $0.4$-near to $2$.
$endgroup$
– janmarqz
Jun 16 '15 at 18:16
add a comment |
$begingroup$
$|sqrt{x}- 2|< 0.4$ is the same as $-0.4< sqrt{x}- 2< 0.4$
Since $sqrt{x}$ is an increasing function, it is sufficient to look at the endpoints of the interval, $-0.4= sqrt{x}- 2$ and $sqrt{x}- 2|= 0.4$.
For $-0.4= sqrt{x}- 2$ we have immediately $sqrt{x}= -0.4+ 2= 1.6$. Squaring both sides, $x= 2.56$. For $sqrt{x}- 2= 0.4$, $sqrt{x}= 2+ 0.4= 2.4$ and $x= 5.76$.
That is, $|sqrt{x}- 4|< 0.4$ is true as long as $2.56< x< 5.76$.
$2.56= 4- 1.44$ and $5.76= 4+ 1.74$. The smaller of those is 1.44 so $|x- 4|< 1.44$ will fit both.
answered Jan 20 at 12:47
user247327user247327
11.2k1515
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take $delta=4-2.56$, so each element in the interval $]4-delta,4+delta[$ is taken, by $f$, into the $0.4$-neighbourhood of $2$.
answered Jun 16 '15 at 17:54
janmarqzjanmarqz
6,20241630
$endgroup$
$begingroup$
Am I trying to find a number $delta$ that is close to 4, so the limit is brought close to a $f(x)$ value that is less than 0.4? I apologize in advance is this question doesn't make much sense, this concept has been really hard for me to wrap my head around.
$endgroup$
– matryoshka
Jun 16 '15 at 18:11
$begingroup$
you are searching for $delta$ to make a $delta$-neighbourhood of $4$ such that for each point in it, they are carried $0.4$-near to $2$.
$endgroup$
– janmarqz
Jun 16 '15 at 18:16
add a comment |
$begingroup$
Take $delta=4-2.56$, so each element in the interval $]4-delta,4+delta[$ is taken, by $f$, into the $0.4$-neighbourhood of $2$.
answered Jun 16 '15 at 17:54
janmarqzjanmarqz
6,20241630
$endgroup$
$begingroup$
Am I trying to find a number $delta$ that is close to 4, so the limit is brought close to a $f(x)$ value that is less than 0.4? I apologize in advance is this question doesn't make much sense, this concept has been really hard for me to wrap my head around.
$endgroup$
– matryoshka
Jun 16 '15 at 18:11
$begingroup$
you are searching for $delta$ to make a $delta$-neighbourhood of $4$ such that for each point in it, they are carried $0.4$-near to $2$.
$endgroup$
– janmarqz
Jun 16 '15 at 18:16
add a comment |
$begingroup$
Take $delta=4-2.56$, so each element in the interval $]4-delta,4+delta[$ is taken, by $f$, into the $0.4$-neighbourhood of $2$.
answered Jun 16 '15 at 17:54
janmarqzjanmarqz
6,20241630
$endgroup$
Take $delta=4-2.56$, so each element in the interval $]4-delta,4+delta[$ is taken, by $f$, into the $0.4$-neighbourhood of $2$.
answered Jun 16 '15 at 17:54
janmarqzjanmarqz
6,20241630
answered Jun 16 '15 at 17:54
janmarqzjanmarqz
6,20241630
answered Jun 16 '15 at 17:54
janmarqzjanmarqz
6,20241630
answered Jun 16 '15 at 17:54
janmarqzjanmarqz
6,20241630
6,20241630
$begingroup$
Am I trying to find a number $delta$ that is close to 4, so the limit is brought close to a $f(x)$ value that is less than 0.4? I apologize in advance is this question doesn't make much sense, this concept has been really hard for me to wrap my head around.
$endgroup$
– matryoshka
Jun 16 '15 at 18:11
$begingroup$
you are searching for $delta$ to make a $delta$-neighbourhood of $4$ such that for each point in it, they are carried $0.4$-near to $2$.
$endgroup$
– janmarqz
Jun 16 '15 at 18:16
add a comment |
$begingroup$
Am I trying to find a number $delta$ that is close to 4, so the limit is brought close to a $f(x)$ value that is less than 0.4? I apologize in advance is this question doesn't make much sense, this concept has been really hard for me to wrap my head around.
$endgroup$
– matryoshka
Jun 16 '15 at 18:11
$begingroup$
you are searching for $delta$ to make a $delta$-neighbourhood of $4$ such that for each point in it, they are carried $0.4$-near to $2$.
$endgroup$
– janmarqz
Jun 16 '15 at 18:16
$begingroup$
Am I trying to find a number $delta$ that is close to 4, so the limit is brought close to a $f(x)$ value that is less than 0.4? I apologize in advance is this question doesn't make much sense, this concept has been really hard for me to wrap my head around.
$endgroup$
– matryoshka
Jun 16 '15 at 18:11
$begingroup$
Am I trying to find a number $delta$ that is close to 4, so the limit is brought close to a $f(x)$ value that is less than 0.4? I apologize in advance is this question doesn't make much sense, this concept has been really hard for me to wrap my head around.
$endgroup$
– matryoshka
Jun 16 '15 at 18:11
$begingroup$
you are searching for $delta$ to make a $delta$-neighbourhood of $4$ such that for each point in it, they are carried $0.4$-near to $2$.
$endgroup$
– janmarqz
Jun 16 '15 at 18:16
$begingroup$
you are searching for $delta$ to make a $delta$-neighbourhood of $4$ such that for each point in it, they are carried $0.4$-near to $2$.
$endgroup$
– janmarqz
Jun 16 '15 at 18:16
add a comment |
$begingroup$
$|sqrt{x}- 2|< 0.4$ is the same as $-0.4< sqrt{x}- 2< 0.4$
Since $sqrt{x}$ is an increasing function, it is sufficient to look at the endpoints of the interval, $-0.4= sqrt{x}- 2$ and $sqrt{x}- 2|= 0.4$.
For $-0.4= sqrt{x}- 2$ we have immediately $sqrt{x}= -0.4+ 2= 1.6$. Squaring both sides, $x= 2.56$. For $sqrt{x}- 2= 0.4$, $sqrt{x}= 2+ 0.4= 2.4$ and $x= 5.76$.
That is, $|sqrt{x}- 4|< 0.4$ is true as long as $2.56< x< 5.76$.
$2.56= 4- 1.44$ and $5.76= 4+ 1.74$. The smaller of those is 1.44 so $|x- 4|< 1.44$ will fit both.
answered Jan 20 at 12:47
user247327user247327
11.2k1515
$endgroup$
add a comment |
$begingroup$
$|sqrt{x}- 2|< 0.4$ is the same as $-0.4< sqrt{x}- 2< 0.4$
Since $sqrt{x}$ is an increasing function, it is sufficient to look at the endpoints of the interval, $-0.4= sqrt{x}- 2$ and $sqrt{x}- 2|= 0.4$.
For $-0.4= sqrt{x}- 2$ we have immediately $sqrt{x}= -0.4+ 2= 1.6$. Squaring both sides, $x= 2.56$. For $sqrt{x}- 2= 0.4$, $sqrt{x}= 2+ 0.4= 2.4$ and $x= 5.76$.
That is, $|sqrt{x}- 4|< 0.4$ is true as long as $2.56< x< 5.76$.
$2.56= 4- 1.44$ and $5.76= 4+ 1.74$. The smaller of those is 1.44 so $|x- 4|< 1.44$ will fit both.
answered Jan 20 at 12:47
user247327user247327
11.2k1515
$endgroup$
add a comment |
$begingroup$
$|sqrt{x}- 2|< 0.4$ is the same as $-0.4< sqrt{x}- 2< 0.4$
Since $sqrt{x}$ is an increasing function, it is sufficient to look at the endpoints of the interval, $-0.4= sqrt{x}- 2$ and $sqrt{x}- 2|= 0.4$.
For $-0.4= sqrt{x}- 2$ we have immediately $sqrt{x}= -0.4+ 2= 1.6$. Squaring both sides, $x= 2.56$. For $sqrt{x}- 2= 0.4$, $sqrt{x}= 2+ 0.4= 2.4$ and $x= 5.76$.
That is, $|sqrt{x}- 4|< 0.4$ is true as long as $2.56< x< 5.76$.
$2.56= 4- 1.44$ and $5.76= 4+ 1.74$. The smaller of those is 1.44 so $|x- 4|< 1.44$ will fit both.
answered Jan 20 at 12:47
user247327user247327
11.2k1515
$endgroup$
$|sqrt{x}- 2|< 0.4$ is the same as $-0.4< sqrt{x}- 2< 0.4$
Since $sqrt{x}$ is an increasing function, it is sufficient to look at the endpoints of the interval, $-0.4= sqrt{x}- 2$ and $sqrt{x}- 2|= 0.4$.
For $-0.4= sqrt{x}- 2$ we have immediately $sqrt{x}= -0.4+ 2= 1.6$. Squaring both sides, $x= 2.56$. For $sqrt{x}- 2= 0.4$, $sqrt{x}= 2+ 0.4= 2.4$ and $x= 5.76$.
That is, $|sqrt{x}- 4|< 0.4$ is true as long as $2.56< x< 5.76$.
$2.56= 4- 1.44$ and $5.76= 4+ 1.74$. The smaller of those is 1.44 so $|x- 4|< 1.44$ will fit both.
answered Jan 20 at 12:47
user247327user247327
11.2k1515
answered Jan 20 at 12:47
user247327user247327
11.2k1515
answered Jan 20 at 12:47
user247327user247327
11.2k1515
answered Jan 20 at 12:47
user247327user247327
11.2k1515
11.2k1515
add a comment |
add a comment |
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