Limits if polar coordinates (conceptual explanation)

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SUMMARY

The limit of the function (xy)/SQRT[x^2 + y^2] as (x,y) approaches (0,0) is confirmed to be zero using polar coordinates. By substituting x = rcosΘ and y = rsinΘ, the expression simplifies to rcosΘsinΘ, which is bounded by -r and r. The significance of the inequality rcosΘsinΘ ≤ r arises from the application of the Squeeze Theorem, confirming that the limit exists and equals zero as r approaches zero.

PREREQUISITES
  • Understanding of polar coordinates and their applications in calculus
  • Familiarity with limits and continuity in multivariable calculus
  • Knowledge of the Squeeze Theorem and its implications
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the Squeeze Theorem in detail to understand its applications in limit proofs
  • Explore polar coordinate transformations in multivariable calculus
  • Learn about the implications of limits in higher dimensions
  • Investigate other examples of limits using polar coordinates for deeper comprehension
USEFUL FOR

Students studying calculus, particularly those focusing on multivariable limits, and educators seeking to explain the application of polar coordinates in limit evaluations.

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1. Homework Statement [/b]

Suppose the lim(x,y) →(0,0) (xy)/SQRT[x^2 + y^2] if it exists

find the limit.

The Attempt at a Solution



x = rcosΘ
y = r sinΘ
r = SQRT[x^2 + y^2]

∴ limr → 0 (r2cosΘrsinΘ)/ r = rcosΘsinΘ [itex]\leq r[/itex]

and so -r [itex]\leq(xy)/SQRT[x^2 + y^2][/itex] [itex]\leq r[/itex]

...
...

I can understand the limit goes to zero because algerbraic multiplication and the sandwhich theorem tells me so. However, the highlighted part in red confuses me. What is the significance of
rcosΘsinΘ [itex]\leqr[/itex]?
 
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It was to generate the inequality - they used cosΘrsinΘ ≤ 1.
 

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