# Math Help - Hard Limit

1. ## Hard Limit

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

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

2. $\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:

$y = x^2 + 2
$

$\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:

$y = x^3 + 2$

$
\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
$\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:

$y = x^2 + 2
$

$\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:

$y = x^3 + 2$

$
\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:

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

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

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

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 $6x^3 \sqrt {x^2 } = 6x^3 \left| x \right|$.
As for 'sign' it is usually written $\left\{ {{\mathop{\rm sgn}} (x) = \begin{array}{ll}
{1\,,\,x > 0} \\
{0\,,\,x = 0} \\
{ - 1\,,\,x < 0} \\
\end{array}} \right.$

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:

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

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:

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

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

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

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

Then, for $y=x^3 + 2$, we get $\frac{6(x^3)^{3/2}}{x^6 + 2x^4}$ and according to my calculator, as $x\to 0$, 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:

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

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

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

Function calculator