MHB What does this Limit mean geometrically?

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SUMMARY

The discussion centers on evaluating the limit $\lim_{(x,y) \to (0,0)} \frac{\cos x-1-\frac{x^2}{2}}{x^4+y^4}$ and understanding its geometric implications. The participants explore the limit of the function $g(x,y)=\frac{xy}{x^2+y^2}$, concluding that its dependence on the parameter $m$ indicates that the limit does not exist. Additionally, they analyze the limit $\lim_{(x,y) \to (0,0)} \frac{\sin{(2x)}-2x+y}{x^3+y}$, noting that differing limits along the axes confirm non-existence. The discussion emphasizes the importance of path independence in determining limit existence.

PREREQUISITES
  • Understanding of multivariable limits in calculus
  • Familiarity with the concept of path dependence in limits
  • Knowledge of Taylor series expansion, particularly for $\cos x$
  • Basic proficiency in evaluating limits involving two variables
NEXT STEPS
  • Study the concept of path independence in multivariable calculus
  • Learn about the epsilon-delta definition of limits in multiple dimensions
  • Explore Taylor series expansions for functions like $\sin x$ and $\cos x$
  • Investigate the implications of limits existing along different paths
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Students and educators in calculus, mathematicians analyzing multivariable functions, and anyone interested in the geometric interpretation of limits in higher dimensions.

evinda
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Hello! (Wave)

I want to find the following limit, if it exists.

$\lim_{(x,y) \to (0,0)} \frac{\cos x-1-\frac{x^2}{2}}{x^4+y^4}$ If we say : let $(x,y) \to (0,0)$ along the line $y=0$ , what exactly does it mean geometrically?

Also, if we want to check whether the limit $\lim_{(x,y) \to (0,0)} g(x,y) $ exists where $g(x,y)=\frac{xy}{x^2+y^2}$, we pick $y=mx$ and since $g(x,mx)$ depends on $m$ we deduce that the limit does not exist.

If $g(x,mx)$ would be independent on $m$, would that mean that the limit exists? If so, why?
 
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evinda said:
Hello! (Wave)

I want to find the following limit, if it exists.

$\lim_{(x,y) \to (0,0)} \frac{\cos x-1-\frac{x^2}{2}}{x^4+y^4}$ If we say : let $(x,y) \to (0,0)$ along the line $y=0$ , what exactly does it mean geometrically?

Also, if we want to check whether the limit $\lim_{(x,y) \to (0,0)} g(x,y) $ exists where $g(x,y)=\frac{xy}{x^2+y^2}$, we pick $y=mx$ and since $g(x,mx)$ depends on $m$ we deduce that the limit does not exist.

If $g(x,mx)$ would be independent on $m$, would that mean that the limit exists? If so, why?

What do you mean by "independent on m"?
 
Prove It said:
What do you mean by "independent on m"?

I mean that the result of the limit isn't a function of $m$...
 
Also, if we want to check if the limit $\lim_{(x,y) \to (0,0)} \frac{\sin{(2x)}-2x+y}{x^3+y}$ exists, we can consider the limit along the line $y=0$ and the limit along $x=0$ and we will see that they are not equal. Is this the only way to see that the limit does not exist?
 
Relativistic Momentum, Mass, and Energy Momentum and mass (...), the classic equations for conserving momentum and energy are not adequate for the analysis of high-speed collisions. (...) The momentum of a particle moving with velocity ##v## is given by $$p=\cfrac{mv}{\sqrt{1-(v^2/c^2)}}\qquad{R-10}$$ ENERGY In relativistic mechanics, as in classic mechanics, the net force on a particle is equal to the time rate of change of the momentum of the particle. Considering one-dimensional...

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