Two-variable limit of $lim_{(x,y)to(0,0)}frac{sin(x^4+y^4)}{x^2+y^2}$
$$lim_{(x,y)to(0,0)}frac{sin(x^4+y^4)}{x^2+y^2}$$
I tried to bound it with $frac{sin((x^2+y^2)^2)}{x^2+y^2}$ and using polar coordinates with $x = rcostheta$ and $y = rsintheta$, but neither of the approaches provided any results. I know that the limit exists and is equal to 0, so tricks with different paths won't work. Should I use the squeeze theorem, or is there another solution?
limits multivariable-calculus
edited yesterday
Lorenzo B.
1,8302520
asked Oct 8 '17 at 8:26
Joald
393314
add a comment |
$$lim_{(x,y)to(0,0)}frac{sin(x^4+y^4)}{x^2+y^2}$$
I tried to bound it with $frac{sin((x^2+y^2)^2)}{x^2+y^2}$ and using polar coordinates with $x = rcostheta$ and $y = rsintheta$, but neither of the approaches provided any results. I know that the limit exists and is equal to 0, so tricks with different paths won't work. Should I use the squeeze theorem, or is there another solution?
limits multivariable-calculus
edited yesterday
Lorenzo B.
1,8302520
asked Oct 8 '17 at 8:26
Joald
393314
add a comment |
$$lim_{(x,y)to(0,0)}frac{sin(x^4+y^4)}{x^2+y^2}$$
I tried to bound it with $frac{sin((x^2+y^2)^2)}{x^2+y^2}$ and using polar coordinates with $x = rcostheta$ and $y = rsintheta$, but neither of the approaches provided any results. I know that the limit exists and is equal to 0, so tricks with different paths won't work. Should I use the squeeze theorem, or is there another solution?
limits multivariable-calculus
edited yesterday
Lorenzo B.
1,8302520
asked Oct 8 '17 at 8:26
Joald
393314
$$lim_{(x,y)to(0,0)}frac{sin(x^4+y^4)}{x^2+y^2}$$
I tried to bound it with $frac{sin((x^2+y^2)^2)}{x^2+y^2}$ and using polar coordinates with $x = rcostheta$ and $y = rsintheta$, but neither of the approaches provided any results. I know that the limit exists and is equal to 0, so tricks with different paths won't work. Should I use the squeeze theorem, or is there another solution?
limits multivariable-calculus
limits multivariable-calculus
edited yesterday
Lorenzo B.
1,8302520
asked Oct 8 '17 at 8:26
Joald
393314
edited yesterday
Lorenzo B.
1,8302520
asked Oct 8 '17 at 8:26
Joald
393314
edited yesterday
Lorenzo B.
1,8302520
edited yesterday
Lorenzo B.
1,8302520
edited yesterday
Lorenzo B.
1,8302520
1,8302520
asked Oct 8 '17 at 8:26
Joald
393314
asked Oct 8 '17 at 8:26
Joald
393314
asked Oct 8 '17 at 8:26
Joald
393314
393314
add a comment |
add a comment |
1 Answer
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Use $|sin t|leq |t|$ then
$$Big|frac{sin(x^4+y^4)}{x^2+y^2}Big|leqfrac{x^4+y^4}{x^2+y^2}leqfrac{x^4+y^4+2x^2y^2}{x^2+y^2}= x^2+y^2$$
edited Oct 8 '17 at 8:35
Andrei
10.9k21025
answered Oct 8 '17 at 8:30
Nosrati
26.4k62353
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Use $|sin t|leq |t|$ then
$$Big|frac{sin(x^4+y^4)}{x^2+y^2}Big|leqfrac{x^4+y^4}{x^2+y^2}leqfrac{x^4+y^4+2x^2y^2}{x^2+y^2}= x^2+y^2$$
edited Oct 8 '17 at 8:35
Andrei
10.9k21025
answered Oct 8 '17 at 8:30
Nosrati
26.4k62353
add a comment |
Use $|sin t|leq |t|$ then
$$Big|frac{sin(x^4+y^4)}{x^2+y^2}Big|leqfrac{x^4+y^4}{x^2+y^2}leqfrac{x^4+y^4+2x^2y^2}{x^2+y^2}= x^2+y^2$$
edited Oct 8 '17 at 8:35
Andrei
10.9k21025
answered Oct 8 '17 at 8:30
Nosrati
26.4k62353
add a comment |
Use $|sin t|leq |t|$ then
$$Big|frac{sin(x^4+y^4)}{x^2+y^2}Big|leqfrac{x^4+y^4}{x^2+y^2}leqfrac{x^4+y^4+2x^2y^2}{x^2+y^2}= x^2+y^2$$
edited Oct 8 '17 at 8:35
Andrei
10.9k21025
answered Oct 8 '17 at 8:30
Nosrati
26.4k62353
Use $|sin t|leq |t|$ then
$$Big|frac{sin(x^4+y^4)}{x^2+y^2}Big|leqfrac{x^4+y^4}{x^2+y^2}leqfrac{x^4+y^4+2x^2y^2}{x^2+y^2}= x^2+y^2$$
edited Oct 8 '17 at 8:35
Andrei
10.9k21025
answered Oct 8 '17 at 8:30
Nosrati
26.4k62353
edited Oct 8 '17 at 8:35
Andrei
10.9k21025
edited Oct 8 '17 at 8:35
Andrei
10.9k21025
edited Oct 8 '17 at 8:35
Andrei
10.9k21025
10.9k21025
answered Oct 8 '17 at 8:30
Nosrati
26.4k62353
answered Oct 8 '17 at 8:30
Nosrati
26.4k62353
answered Oct 8 '17 at 8:30
Nosrati
26.4k62353
26.4k62353
add a comment |
add a comment |
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