find the limit of the function, if it exists as x and y approach zero
[y^2sin^2(x)]/[x^4+y^4]
[xy]/[x^2+y^2]^(1/2)
Much appreciated!
[y^2sin^2(x)]/[x^4+y^4]
[xy]/[x^2+y^2]^(1/2)
Much appreciated!
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1) Try different paths to the origin.
x = 0 ==> lim(y→0) 0/x^4 = 0.
y = x ==> lim(x→0) y^2 sin^2(x) / (2y^4) = (1/2) lim(x→0) [sin(x)/x]^2 = 1/2 * 1^2 = 1/2.
Since we get different limits on different paths, the multivariable limit does not exist.
2) Use polar coordinates.
==> lim(r→0+) (r^2 cos t sin t) / r = lim(r→0) r cos t sin t.
Since -1 ≤ cos t sin t ≤ 1 for all t, we have
-r ≤ r cos t sin t ≤ r cos t sin t for all t.
Since lim(r→0+) -r = 0 = lim(r→0+) r, we conclude by the Squeeze Theorem that
lim(r→0+) r sin t cos t = 0.
I hope this helps!
x = 0 ==> lim(y→0) 0/x^4 = 0.
y = x ==> lim(x→0) y^2 sin^2(x) / (2y^4) = (1/2) lim(x→0) [sin(x)/x]^2 = 1/2 * 1^2 = 1/2.
Since we get different limits on different paths, the multivariable limit does not exist.
2) Use polar coordinates.
==> lim(r→0+) (r^2 cos t sin t) / r = lim(r→0) r cos t sin t.
Since -1 ≤ cos t sin t ≤ 1 for all t, we have
-r ≤ r cos t sin t ≤ r cos t sin t for all t.
Since lim(r→0+) -r = 0 = lim(r→0+) r, we conclude by the Squeeze Theorem that
lim(r→0+) r sin t cos t = 0.
I hope this helps!
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x^4 +x^4 = 2x^4; that's wehere the 2 came from in the first question.
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