Two-variable limit of $lim_{(x,y)to(0,0)}frac{sin(x^4+y^4)}{x^2+y^2}$
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$$lim_{(x,y)to(0,0)}frac{sin(x^4+y^4)}{x^2+y^2}$$
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
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add a comment |
$begingroup$
$$lim_{(x,y)to(0,0)}frac{sin(x^4+y^4)}{x^2+y^2}$$
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
$endgroup$
add a comment |
$begingroup$
$$lim_{(x,y)to(0,0)}frac{sin(x^4+y^4)}{x^2+y^2}$$
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
$endgroup$
$$lim_{(x,y)to(0,0)}frac{sin(x^4+y^4)}{x^2+y^2}$$
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 Dec 24 '18 at 22:55
Lorenzo B.
1,8622520
1,8622520
asked Oct 8 '17 at 8:26
JoaldJoald
393314
393314
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1 Answer
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$begingroup$
Use $|sin t|leq |t|$ then
$$Big|frac{sin(x^4+y^4)}{x^2+y^2}Big|leqfrac{x^4+y^4}{x^2+y^2}leqfrac{x^4+y^4+2x^2y^2}{x^2+y^2}= x^2+y^2$$
$endgroup$
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1 Answer
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active
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1 Answer
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active
oldest
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active
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active
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votes
$begingroup$
Use $|sin t|leq |t|$ then
$$Big|frac{sin(x^4+y^4)}{x^2+y^2}Big|leqfrac{x^4+y^4}{x^2+y^2}leqfrac{x^4+y^4+2x^2y^2}{x^2+y^2}= x^2+y^2$$
$endgroup$
add a comment |
$begingroup$
Use $|sin t|leq |t|$ then
$$Big|frac{sin(x^4+y^4)}{x^2+y^2}Big|leqfrac{x^4+y^4}{x^2+y^2}leqfrac{x^4+y^4+2x^2y^2}{x^2+y^2}= x^2+y^2$$
$endgroup$
add a comment |
$begingroup$
Use $|sin t|leq |t|$ then
$$Big|frac{sin(x^4+y^4)}{x^2+y^2}Big|leqfrac{x^4+y^4}{x^2+y^2}leqfrac{x^4+y^4+2x^2y^2}{x^2+y^2}= x^2+y^2$$
$endgroup$
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
13.1k21230
13.1k21230
answered Oct 8 '17 at 8:30
NosratiNosrati
26.6k62354
26.6k62354
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