Multivariable limit evaluation

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SUMMARY

The discussion focuses on evaluating the multivariable limit of the expression lim (x,y) → (0,0) [sin(2x) - 2x + y] / [x^3 + y]. Multiple approaches, including paths along the x-axis, y-axis, and y = x, yield the indeterminate form 0/0. Participants recommend using L'Hôpital's Rule, which is applicable for single-variable limits, and suggest evaluating the limit along the line y = mx or the curve y = m(x^3) to resolve the indeterminate form.

PREREQUISITES
  • Understanding of multivariable limits
  • Familiarity with L'Hôpital's Rule
  • Knowledge of trigonometric functions and their limits
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Learn how to apply L'Hôpital's Rule in multivariable contexts
  • Study techniques for approaching limits along different paths
  • Explore the concept of continuity and differentiability in multivariable calculus
  • Investigate the behavior of limits involving trigonometric functions
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Students and educators in calculus, particularly those focusing on multivariable limits, as well as mathematicians seeking to deepen their understanding of limit evaluation techniques.

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Homework Statement




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

Homework Equations





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'hospital's rule? If so I'm not sure how to use it here...
 
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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,

y=mx

It'll follow quite naturally from this.
 
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?
 
For (2 cos(2x) -2)/3x2 , use L'Hôpital's rule again, then again.
 
Try approaching (0,0) along the curve, y = m(x3)
 

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