MHB Evaluating limit limit of multivariable function

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The limit of the multivariable function as (x, y) approaches (0, 0) is evaluated using different paths, leading to conflicting results. Approaching along the x-axis yields a limit of 0, while approaching along the y-axis also results in 0. However, applying L'Hôpital's rule suggests that the limit can evaluate to infinity, indicating that the limit does not exist. The discussion highlights the importance of understanding the conditions under which L'Hôpital's rule applies. Ultimately, the limit's existence is questioned due to these differing outcomes.
tmt1
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I have

$$\lim_{{(x, y)}\to{(0, 0)}} \frac{x}{x^2 + y^2}$$

We can approach the limit on the x-axis, so the values of $x$ will change and the values of $y$ will stay :

$$\lim_{{x}\to{0}} \frac{x}{x^2}$$

I suppose I can take hospital's rule and get

$$\lim_{{x}\to{0}} \frac{x}{x^2}$$

$$\lim_{{x}\to{0}} \frac{1}{2x}$$

and

$$\lim_{{x}\to{0}} \frac{0}{2}$$

so the limit is 0.

Then we can approach the limit on the y-axis, so the values of $y$ will change and the values of $x$ will change.

$$\lim_{{y}\to{0}} \frac{0}{0 + y^2}$$

Which is 0. Because no matter what the value of y, the result will be zero.

Therefore, the limit of the function exists and it is 0.

However, in the text it says that the limit of this function does not exist.
 
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tmt said:
I suppose I can take hospital's rule and get

$$\lim_{{x}\to{0}} \frac{x}{x^2}$$

$$\lim_{{x}\to{0}} \frac{1}{2x}$$

and

$$\lim_{{x}\to{0}} \frac{0}{2}$$

so the limit is 0.
What? Do you have to use L'Hopital's rule?
 
Evgeny.Makarov said:
What? Do you have to use L'Hopital's rule?

Ah yes, you're right, the limit actually evaluates to infinity if I don't use L'Hopital's rule.

So, I suppose that means the limit doesn't exist in that case.
 
It's instructive to find the condition of the L'Hopital's rule theorem that does not hold for $1/x$.
 

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