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Multivariable limit evaluation

  • Thread starter Kuma
  • Start date
134
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1. Homework Statement


lim
(x,y) --> (0,0) [sin(2x) -2x + y]/[x^3+y]

2. Homework Equations



3. The Attempt at a Solution

so I tried approaching from the x axis and I got

y=0 lim x-> 0 [sin(2x) -x + 0]/[x^3 + 0] which gives me 0/0

and from the y axis

y=0 lim y ->0 which gives me lim y-> 0 y/y = 0/0

if I try approaching from a line such as y = x I also get 0/0

what do I do now? L'hopitals rule? If so I'm not sure how to use it here...
 
I think you'll find that, for any line you choose to approach the origin with, you'll get a single variable limit of 0/0 form. You may apply L'Hospital's Rule for this. Remember that L'Hospital's Rule only works for single variable limits.

If you are having difficulty getting differing limits, try approaching the origin via the line,

[tex]y=mx[/tex]

It'll follow quite naturally from this.
 
134
0
okay so then using l hopitals rule for this, since it only applies for single variable limits, if I choose y=0 and evaluate it at

lim x-> 0 I get sin(2x) - 2x/x^3 then using l hopitals rule which is

lim x->a f'(x)/g'(x) I get

2 cos(2x) -2/3x^2, evaluating that at x=0 I still get 0/0...

so what now?
 

SammyS

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For (2 cos(2x) -2)/3x2 , use L'Hôpital's rule again, then again.
 

SammyS

Staff Emeritus
Science Advisor
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Gold Member
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Try approaching (0,0) along the curve, y = m(x3)
 

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