Multivariable limit evaluation

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Homework Help Overview

The problem involves evaluating a multivariable limit as (x,y) approaches (0,0) for the expression [sin(2x) - 2x + y]/[x^3 + y].

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss various approaches to the limit, including evaluating along the x-axis and y-axis, which both yield a 0/0 indeterminate form. There is mention of using L'Hospital's Rule and the suggestion to approach along the line y=mx.

Discussion Status

The discussion is ongoing, with participants exploring different paths to evaluate the limit. Some have suggested applying L'Hospital's Rule multiple times, while others propose alternative approaches such as approaching along specific curves.

Contextual Notes

There is a recognition that L'Hospital's Rule applies only to single-variable limits, and participants are considering how to adapt their approaches accordingly. The repeated occurrence of the 0/0 form is noted as a challenge in the evaluation process.

Kuma
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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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