Mixing
Epsilon-Delta proof of limit as $(x,y)to 0,0)$ of $sin(x^2+y^2)=0$
$begingroup$
I am trying to come up with an epsilon delta proof of
$$lim_{(x,y)to (0,0)}sin(x^2+y^2)=0.$$
I know I have to find the form square root of $x^2+y^2 < delta$, but the only thing I know is that the absolute value of $sin (x^2+y^2) <= 1$
Thanks!
calculus limits multivariable-calculus epsilon-delta
edited Jan 31 at 22:35
Antonios-Alexandros Robotis
10.6k41741
asked Jan 31 at 22:28
Edgar LozanoEdgar Lozano
362
$endgroup$
add a comment |
$begingroup$
I am trying to come up with an epsilon delta proof of
$$lim_{(x,y)to (0,0)}sin(x^2+y^2)=0.$$
I know I have to find the form square root of $x^2+y^2 < delta$, but the only thing I know is that the absolute value of $sin (x^2+y^2) <= 1$
Thanks!
calculus limits multivariable-calculus epsilon-delta
edited Jan 31 at 22:35
Antonios-Alexandros Robotis
10.6k41741
asked Jan 31 at 22:28
Edgar LozanoEdgar Lozano
362
$endgroup$
$begingroup$
$xto 0, yto 0$ imply $x^2+y^2to 0$, while $sin(u)$ is continuous and $sin(0)=0$ should be enough.
$endgroup$
– herb steinberg
Jan 31 at 22:33
$begingroup$
$sin(x^2 + y^2) le x^2 + y^2 = delta^2.$ Let $delta = sqrt{epsilon}$
$endgroup$
– Doug M
Jan 31 at 22:40
add a comment |
$begingroup$
I am trying to come up with an epsilon delta proof of
$$lim_{(x,y)to (0,0)}sin(x^2+y^2)=0.$$
I know I have to find the form square root of $x^2+y^2 < delta$, but the only thing I know is that the absolute value of $sin (x^2+y^2) <= 1$
Thanks!
calculus limits multivariable-calculus epsilon-delta
edited Jan 31 at 22:35
Antonios-Alexandros Robotis
10.6k41741
asked Jan 31 at 22:28
Edgar LozanoEdgar Lozano
362
$endgroup$
I am trying to come up with an epsilon delta proof of
$$lim_{(x,y)to (0,0)}sin(x^2+y^2)=0.$$
I know I have to find the form square root of $x^2+y^2 < delta$, but the only thing I know is that the absolute value of $sin (x^2+y^2) <= 1$
Thanks!
calculus limits multivariable-calculus epsilon-delta
calculus limits multivariable-calculus epsilon-delta
edited Jan 31 at 22:35
Antonios-Alexandros Robotis
10.6k41741
asked Jan 31 at 22:28
Edgar LozanoEdgar Lozano
362
edited Jan 31 at 22:35
Antonios-Alexandros Robotis
10.6k41741
asked Jan 31 at 22:28
Edgar LozanoEdgar Lozano
362
edited Jan 31 at 22:35
Antonios-Alexandros Robotis
10.6k41741
edited Jan 31 at 22:35
Antonios-Alexandros Robotis
10.6k41741
edited Jan 31 at 22:35
Antonios-Alexandros Robotis
10.6k41741
10.6k41741
asked Jan 31 at 22:28
Edgar LozanoEdgar Lozano
362
asked Jan 31 at 22:28
Edgar LozanoEdgar Lozano
362
asked Jan 31 at 22:28
Edgar LozanoEdgar Lozano
362
362
$begingroup$
$xto 0, yto 0$ imply $x^2+y^2to 0$, while $sin(u)$ is continuous and $sin(0)=0$ should be enough.
$endgroup$
– herb steinberg
Jan 31 at 22:33
$begingroup$
$sin(x^2 + y^2) le x^2 + y^2 = delta^2.$ Let $delta = sqrt{epsilon}$
$endgroup$
– Doug M
Jan 31 at 22:40
add a comment |
$begingroup$
$xto 0, yto 0$ imply $x^2+y^2to 0$, while $sin(u)$ is continuous and $sin(0)=0$ should be enough.
$endgroup$
– herb steinberg
Jan 31 at 22:33
$begingroup$
$sin(x^2 + y^2) le x^2 + y^2 = delta^2.$ Let $delta = sqrt{epsilon}$
$endgroup$
– Doug M
Jan 31 at 22:40
$begingroup$
$xto 0, yto 0$ imply $x^2+y^2to 0$, while $sin(u)$ is continuous and $sin(0)=0$ should be enough.
$endgroup$
– herb steinberg
Jan 31 at 22:33
$begingroup$
$xto 0, yto 0$ imply $x^2+y^2to 0$, while $sin(u)$ is continuous and $sin(0)=0$ should be enough.
$endgroup$
– herb steinberg
Jan 31 at 22:33
$begingroup$
$sin(x^2 + y^2) le x^2 + y^2 = delta^2.$ Let $delta = sqrt{epsilon}$
$endgroup$
– Doug M
Jan 31 at 22:40
$begingroup$
$sin(x^2 + y^2) le x^2 + y^2 = delta^2.$ Let $delta = sqrt{epsilon}$
$endgroup$
– Doug M
Jan 31 at 22:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can make use of the fact that for $0 < z < 1$, $0 < sin z < z$.
For $0 < x, y < 1$, then, $0 < x^2 + y^2 < 1$, and hence $0 < sin (x^2 + y^2) < x^2 + y^2$.
From this point, we can then use the squeeze theorem and state that since $x^2 + y^2 rightarrow 0$ in the given limit, $sin (x^2 + y^2)$ must also approach $0$. You can dress that up in $epsilon-delta$ notation as required.
answered Jan 31 at 22:40
ConManConMan
7,9771324
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
We can make use of the fact that for $0 < z < 1$, $0 < sin z < z$.
For $0 < x, y < 1$, then, $0 < x^2 + y^2 < 1$, and hence $0 < sin (x^2 + y^2) < x^2 + y^2$.
From this point, we can then use the squeeze theorem and state that since $x^2 + y^2 rightarrow 0$ in the given limit, $sin (x^2 + y^2)$ must also approach $0$. You can dress that up in $epsilon-delta$ notation as required.
answered Jan 31 at 22:40
ConManConMan
7,9771324
$endgroup$
add a comment |
$begingroup$
We can make use of the fact that for $0 < z < 1$, $0 < sin z < z$.
For $0 < x, y < 1$, then, $0 < x^2 + y^2 < 1$, and hence $0 < sin (x^2 + y^2) < x^2 + y^2$.
From this point, we can then use the squeeze theorem and state that since $x^2 + y^2 rightarrow 0$ in the given limit, $sin (x^2 + y^2)$ must also approach $0$. You can dress that up in $epsilon-delta$ notation as required.
answered Jan 31 at 22:40
ConManConMan
7,9771324
$endgroup$
add a comment |
$begingroup$
We can make use of the fact that for $0 < z < 1$, $0 < sin z < z$.
For $0 < x, y < 1$, then, $0 < x^2 + y^2 < 1$, and hence $0 < sin (x^2 + y^2) < x^2 + y^2$.
From this point, we can then use the squeeze theorem and state that since $x^2 + y^2 rightarrow 0$ in the given limit, $sin (x^2 + y^2)$ must also approach $0$. You can dress that up in $epsilon-delta$ notation as required.
answered Jan 31 at 22:40
ConManConMan
7,9771324
$endgroup$
We can make use of the fact that for $0 < z < 1$, $0 < sin z < z$.
For $0 < x, y < 1$, then, $0 < x^2 + y^2 < 1$, and hence $0 < sin (x^2 + y^2) < x^2 + y^2$.
From this point, we can then use the squeeze theorem and state that since $x^2 + y^2 rightarrow 0$ in the given limit, $sin (x^2 + y^2)$ must also approach $0$. You can dress that up in $epsilon-delta$ notation as required.
answered Jan 31 at 22:40
ConManConMan
7,9771324
answered Jan 31 at 22:40
ConManConMan
7,9771324
answered Jan 31 at 22:40
ConManConMan
7,9771324
answered Jan 31 at 22:40
ConManConMan
7,9771324
7,9771324
add a comment |
add a comment |
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$begingroup$
$xto 0, yto 0$ imply $x^2+y^2to 0$, while $sin(u)$ is continuous and $sin(0)=0$ should be enough.
$endgroup$
– herb steinberg
Jan 31 at 22:33
$begingroup$
$sin(x^2 + y^2) le x^2 + y^2 = delta^2.$ Let $delta = sqrt{epsilon}$
$endgroup$
– Doug M
Jan 31 at 22:40