Proving multivariable limit using epsilon-delta definition
$begingroup$
Let $displaystyle g(x, y) = frac{sin^2(x - y)}{|x| + |y|}$. Prove
that $lim_{(x, y)to(0,0)} g(x) = 0$.
Any help on this would be appreciated. So far I think that the fact that $|x + y| le |x| + |y|$ and $|sin(x + y)| le |x + y|$ has to be utilized but I am not sure how to do so in the epsilon delta proof.
limits multivariable-calculus
edited Jun 11 '13 at 6:19
Potato
21.4k1189189
$endgroup$
add a comment |
$begingroup$
Let $displaystyle g(x, y) = frac{sin^2(x - y)}{|x| + |y|}$. Prove
that $lim_{(x, y)to(0,0)} g(x) = 0$.
Any help on this would be appreciated. So far I think that the fact that $|x + y| le |x| + |y|$ and $|sin(x + y)| le |x + y|$ has to be utilized but I am not sure how to do so in the epsilon delta proof.
limits multivariable-calculus
edited Jun 11 '13 at 6:19
Potato
21.4k1189189
$endgroup$
add a comment |
$begingroup$
Let $displaystyle g(x, y) = frac{sin^2(x - y)}{|x| + |y|}$. Prove
that $lim_{(x, y)to(0,0)} g(x) = 0$.
Any help on this would be appreciated. So far I think that the fact that $|x + y| le |x| + |y|$ and $|sin(x + y)| le |x + y|$ has to be utilized but I am not sure how to do so in the epsilon delta proof.
limits multivariable-calculus
edited Jun 11 '13 at 6:19
Potato
21.4k1189189
$endgroup$
Let $displaystyle g(x, y) = frac{sin^2(x - y)}{|x| + |y|}$. Prove
that $lim_{(x, y)to(0,0)} g(x) = 0$.
Any help on this would be appreciated. So far I think that the fact that $|x + y| le |x| + |y|$ and $|sin(x + y)| le |x + y|$ has to be utilized but I am not sure how to do so in the epsilon delta proof.
limits multivariable-calculus
limits multivariable-calculus
edited Jun 11 '13 at 6:19
Potato
21.4k1189189
edited Jun 11 '13 at 6:19
Potato
21.4k1189189
edited Jun 11 '13 at 6:19
Potato
21.4k1189189
edited Jun 11 '13 at 6:19
Potato
21.4k1189189
edited Jun 11 '13 at 6:19
Potato
21.4k1189189
21.4k1189189
asked Jun 11 '13 at 6:05
user33969
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2 Answers
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$begingroup$
Remember that $|sin(z)| < z$ for $z ne 0$. So $0 le sin^2(z) < z^2$. Set $z = x-y$. Your denominator satisfies $|z| = |x-y| le |x| + |y|$ by the triangle inequality.
$$0 le lim_{(x,y) to (0,0)} frac{sin^2(x-y)}{|x|+|y|} le lim_{(x,y) to (0,0)} frac{z^2}{|z|} = lim_{(x,y) to (0,0)} |z| = 0$$
You can easily convert this proof, which uses the squeeze theorem, into an $epsilon-delta$ proof by noting the inequalities and setting $delta = epsilon$. For $Vert ( x,y ) Vert < epsilon$, we have $|z| < epsilon$ since $|x-y| le sqrt{x^2 + y^2}$.
answered Jun 11 '13 at 6:17
A.SA.S
7,3881855
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$begingroup$
An idea:
$$|x-y|le|x|+|y|implies frac{sin^2(x-y)}{|x|+|y|}lefrac{sin^2(x-y)}{|x-y|}=|sin(x-y)|frac{|sin(x-y)|}{|x-y|}$$
Now just remember that $,displaystyle{lim_{tto 0}frac{sin t}t=1};$ and stuff...
answered Jun 11 '13 at 6:19
DonAntonioDonAntonio
178k1492227
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2 Answers
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
Remember that $|sin(z)| < z$ for $z ne 0$. So $0 le sin^2(z) < z^2$. Set $z = x-y$. Your denominator satisfies $|z| = |x-y| le |x| + |y|$ by the triangle inequality.
$$0 le lim_{(x,y) to (0,0)} frac{sin^2(x-y)}{|x|+|y|} le lim_{(x,y) to (0,0)} frac{z^2}{|z|} = lim_{(x,y) to (0,0)} |z| = 0$$
You can easily convert this proof, which uses the squeeze theorem, into an $epsilon-delta$ proof by noting the inequalities and setting $delta = epsilon$. For $Vert ( x,y ) Vert < epsilon$, we have $|z| < epsilon$ since $|x-y| le sqrt{x^2 + y^2}$.
answered Jun 11 '13 at 6:17
A.SA.S
7,3881855
$endgroup$
add a comment |
$begingroup$
Remember that $|sin(z)| < z$ for $z ne 0$. So $0 le sin^2(z) < z^2$. Set $z = x-y$. Your denominator satisfies $|z| = |x-y| le |x| + |y|$ by the triangle inequality.
$$0 le lim_{(x,y) to (0,0)} frac{sin^2(x-y)}{|x|+|y|} le lim_{(x,y) to (0,0)} frac{z^2}{|z|} = lim_{(x,y) to (0,0)} |z| = 0$$
You can easily convert this proof, which uses the squeeze theorem, into an $epsilon-delta$ proof by noting the inequalities and setting $delta = epsilon$. For $Vert ( x,y ) Vert < epsilon$, we have $|z| < epsilon$ since $|x-y| le sqrt{x^2 + y^2}$.
answered Jun 11 '13 at 6:17
A.SA.S
7,3881855
$endgroup$
add a comment |
$begingroup$
Remember that $|sin(z)| < z$ for $z ne 0$. So $0 le sin^2(z) < z^2$. Set $z = x-y$. Your denominator satisfies $|z| = |x-y| le |x| + |y|$ by the triangle inequality.
$$0 le lim_{(x,y) to (0,0)} frac{sin^2(x-y)}{|x|+|y|} le lim_{(x,y) to (0,0)} frac{z^2}{|z|} = lim_{(x,y) to (0,0)} |z| = 0$$
You can easily convert this proof, which uses the squeeze theorem, into an $epsilon-delta$ proof by noting the inequalities and setting $delta = epsilon$. For $Vert ( x,y ) Vert < epsilon$, we have $|z| < epsilon$ since $|x-y| le sqrt{x^2 + y^2}$.
answered Jun 11 '13 at 6:17
A.SA.S
7,3881855
$endgroup$
Remember that $|sin(z)| < z$ for $z ne 0$. So $0 le sin^2(z) < z^2$. Set $z = x-y$. Your denominator satisfies $|z| = |x-y| le |x| + |y|$ by the triangle inequality.
$$0 le lim_{(x,y) to (0,0)} frac{sin^2(x-y)}{|x|+|y|} le lim_{(x,y) to (0,0)} frac{z^2}{|z|} = lim_{(x,y) to (0,0)} |z| = 0$$
You can easily convert this proof, which uses the squeeze theorem, into an $epsilon-delta$ proof by noting the inequalities and setting $delta = epsilon$. For $Vert ( x,y ) Vert < epsilon$, we have $|z| < epsilon$ since $|x-y| le sqrt{x^2 + y^2}$.
answered Jun 11 '13 at 6:17
A.SA.S
7,3881855
answered Jun 11 '13 at 6:17
A.SA.S
7,3881855
answered Jun 11 '13 at 6:17
A.SA.S
7,3881855
answered Jun 11 '13 at 6:17
A.SA.S
7,3881855
7,3881855
add a comment |
add a comment |
$begingroup$
An idea:
$$|x-y|le|x|+|y|implies frac{sin^2(x-y)}{|x|+|y|}lefrac{sin^2(x-y)}{|x-y|}=|sin(x-y)|frac{|sin(x-y)|}{|x-y|}$$
Now just remember that $,displaystyle{lim_{tto 0}frac{sin t}t=1};$ and stuff...
answered Jun 11 '13 at 6:19
DonAntonioDonAntonio
178k1492227
$endgroup$
add a comment |
$begingroup$
An idea:
$$|x-y|le|x|+|y|implies frac{sin^2(x-y)}{|x|+|y|}lefrac{sin^2(x-y)}{|x-y|}=|sin(x-y)|frac{|sin(x-y)|}{|x-y|}$$
Now just remember that $,displaystyle{lim_{tto 0}frac{sin t}t=1};$ and stuff...
answered Jun 11 '13 at 6:19
DonAntonioDonAntonio
178k1492227
$endgroup$
add a comment |
$begingroup$
An idea:
$$|x-y|le|x|+|y|implies frac{sin^2(x-y)}{|x|+|y|}lefrac{sin^2(x-y)}{|x-y|}=|sin(x-y)|frac{|sin(x-y)|}{|x-y|}$$
Now just remember that $,displaystyle{lim_{tto 0}frac{sin t}t=1};$ and stuff...
answered Jun 11 '13 at 6:19
DonAntonioDonAntonio
178k1492227
$endgroup$
An idea:
$$|x-y|le|x|+|y|implies frac{sin^2(x-y)}{|x|+|y|}lefrac{sin^2(x-y)}{|x-y|}=|sin(x-y)|frac{|sin(x-y)|}{|x-y|}$$
Now just remember that $,displaystyle{lim_{tto 0}frac{sin t}t=1};$ and stuff...
answered Jun 11 '13 at 6:19
DonAntonioDonAntonio
178k1492227
answered Jun 11 '13 at 6:19
DonAntonioDonAntonio
178k1492227
answered Jun 11 '13 at 6:19
DonAntonioDonAntonio
178k1492227
answered Jun 11 '13 at 6:19
DonAntonioDonAntonio
178k1492227
178k1492227
add a comment |
add a comment |
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