Evaluating limit limit of multivariable function

[tmt1]

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.

 

[Evgeny.Makarov]

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?

 

[tmt1]

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.

 

[Evgeny.Makarov]

It's instructive to find the condition of the L'Hopital's rule theorem that does not hold for $1/x$.