1. Hard Limit

Find the following limit, or show that the limit doesn't exist.

$\displaystyle \lim_{(x,y)\to (0,2)}{\frac{6x^3\sqrt{y-2}}{2x^4 + y^2 - 4y + 4}}$

2. $\displaystyle \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^3 \sqrt {y - 2} }} {{2x^4 + \left( {y - 2} \right)^2 }}$

lets look what's the limit when it is approached along the curve:

$\displaystyle y = x^2 + 2$

$\displaystyle \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^4 }} {{3x^4 }} = 2$

now lets look what's the limit when it is approached along the curve:

$\displaystyle y = x^3 + 2$

$\displaystyle \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^{4\frac{1} {2}} }} {{2x^4 + x^6 }} = \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^{\frac{1} {2}} }} {{2 + x^2 }} = 0$

Since taking different paths toward the same point yields different values for the limit, the limit does not exist.

3. Originally Posted by Peritus
$\displaystyle \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^3 \sqrt {y - 2} }} {{2x^4 + \left( {y - 2} \right)^2 }}$

lets look what's the limit when it is approached along the curve:

$\displaystyle y = x^2 + 2$

$\displaystyle \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^4 }} {{3x^4 }} = 2$

now lets look what's the limit when it is approached along the curve:

$\displaystyle y = x^3 + 2$

$\displaystyle \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^{4\frac{1} {2}} }} {{2x^4 + x^6 }} = \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^{\frac{1} {2}} }} {{2 + x^2 }} = 0$

Since taking different paths toward the same point yields different values for the limit, the limit does not exist.
When plugging x^2 + 2 in for y, we get:

$\displaystyle \frac{6x^3\sqrt{x^2 + 2 -2}}{2x^4+(x^2+2-2)^2}$; how's that equal to $\displaystyle \frac{6x^4}{3x^4}$ .

And then when plugging in x^3 + 2 for y, we get:

$\displaystyle \frac{18}{x^3 + 2x}$. As $\displaystyle x \to 2$, that limit is $\displaystyle \frac{3}{2}$..

EDIT: when plugging in x^3 + 2, I get (6sign(x))/(x^2+2) .. no clue what the "sign" is.

4. Actually you are correct to question that reply.
It should be $\displaystyle 6x^3 \sqrt {x^2 } = 6x^3 \left| x \right|$.
As for 'sign' it is usually written $\displaystyle \left\{ {{\mathop{\rm sgn}} (x) = \begin{array}{ll} {1\,,\,x > 0} \\ {0\,,\,x = 0} \\ { - 1\,,\,x < 0} \\ \end{array}} \right.$

That fits with the absolute value.
Nonetheless, the idea given in the first reply is correct.

5. a little correction: after the first substitution we get:

$\displaystyle 2\frac{{\left| x \right|}} {x}$

when x approaches 0 the limit of this function is undefined.

6. Originally Posted by Peritus
a little correction: after the first substitution we get:

$\displaystyle 2\frac{{\left| x \right|}} {x}$

when x approaches 0 the limit of this function is undefined.
Ok let me know if I'm wrong.

For $\displaystyle y = x^2 +2$, we get:

$\displaystyle \frac{6x|x^3|}{3x^4}$ and as $\displaystyle x\to 0$, that is undefined.

Then, for $\displaystyle y=x^3 + 2$, we get $\displaystyle \frac{6(x^3)^{3/2}}{x^6 + 2x^4}$ and according to my calculator, as $\displaystyle x\to 0$, I'm unable to calculate the limit... (but since we already got one undefined value, the limit doesn't exist?)

7. now as I've already shown you divide the expression by x^4:

$\displaystyle \frac{{6x^{{\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{$2$}}} }} {{x^2 + 2}}$

as you can clearly see when x approaches 0 we get:

$\displaystyle \frac{{6 \cdot 0}} {{0 + 2}} = 0$

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