$$\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^3 + y}$$
I tried the following, but im not sure if this is allowed:
Let $$t = x^3+y$$ $$(x,y)\to (0,0)$$ $$t\to 0$$ $$y=t-x^3$$ Now we will calculate the limit: $$\lim_{(x,t) \to (0,0)}x^2-\frac{x^5}{t}$$ Lets look at the path $t=x^6$ $$\lim_{(x,t) \to (0,0)}x^2-\frac{x^5}{x^6}=-\infty$$ Now lets look at the path $t=x^5$ $$\lim_{(x,t) \to (0,0)}x^2-\frac{x^5}{x^5}=-1$$ We found 2 different limits hence the limit does not exist.
Is this correct? or am I doing something wrong? thanks.
$\endgroup$ 02 Answers
$\begingroup$The function is not continuous at (0,0). Another way to see this is to take the limit along $y = 0 $ and $y = x^6 - x^3$. Your approach is correct.
$\endgroup$ 1 $\begingroup$In multivariable calculus the approach to show that a limit does not exist is exactly the approach you used. Although you used an extreme hard path, in my opinion, you could try to use some simple cases like:
$$ y = 0 $$ $$ x = 0 $$ $$ y = x^2 $$ $$ y = -x $$ $$ y = x $$
If, in these five cases, it converges to the same limit value, it probably - but not necessarily - has a well defined limit. Otherwise it doesn't have one.
As suggested by @mic, this technique only proves that a limit doesn't exist, but cannot prove that it exists. It just suggests a guess for its value. A good way to prove would be to use the delta, epsilon definition of limits.
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