1. ## mulitvariable limit

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

I know can't use L'Hospitals rule anymore and I don't see a way of factoring to clean it up. My intuition tells me the limit does not exist.

I just need help trying to prove it, the question specifically says "do not attempt a precise justification"

2. ## Re: mulitvariable limit

in the same set of questions

$lim_{(x,y) \to (0,0)} \frac{e^{xy}}{x+1}$

seems pretty clear that this goes to one, is using substitution fair in this??

3. ## Re: mulitvariable limit

Let $f:\Bbb{R} \to \Bbb{R}$ be any continuous function with $f(0) = 0$. If you find replace $y = f(x)$ and find the limit as $x \to 0$, and find that the limit changes if you change the function $f$, then the original limit does not exist. This is easiest to do with lines. Note: This method only "proves" when no limit exists. Just because a limit is constant for all linear approaches does not necessarily mean the original limit existed.

So, let's approach using two lines. First, let's approach on the $y$-axis ( $x=0$). Then we are looking at

$\lim_{y \to 0} \dfrac{4(0)^2+3y^2+(0)^3y^3}{(0)^2+y^2+(0)^4y^4} = 3$

Next, let's approach on the $x$-axis ( $y=0$). Then we are looking at

$\lim_{x \to 0} \dfrac{4x^2+3(0)^2+x^3(0)^3}{x^2+(0)^2+x^4(0)^4} = 4$

Since these limits are not equal, you are correct that no limit exists for the first problem. If they had been equal, that would NOT have guaranteed that the limit did exist.

For the second problem, use limit laws:

$\lim_{(x,y) \to (0,0)} \dfrac{e^{xy}}{x+1} = \dfrac{\displaystyle \lim_{(x,y) \to (0,0)} e^{xy}}{\displaystyle \lim_{(x,y) \to (0,0)} x+1} = \dfrac{1}{1} = 1$

Since the denominator was not approaching zero, you have a continuous rational function, so yes, as you suspected, substitution is fair in this one.

4. ## Re: mulitvariable limit

Originally Posted by Jonroberts74
$lim_{(x,y) \to (0,0)} \frac{4x^2+3y^2 +x^3y^3}{x^2+y^2+x^4y^4}$

I know can't use L'Hospitals rule anymore and I don't see a way of factoring to clean it up. My intuition tells me the limit does not exist.

I just need help trying to prove it, the question specifically says "do not attempt a precise justification"
When you have powers of x and y, I tend to like converting to polars...

\displaystyle \begin{align*} \lim_{(x,y) \to (0,0)} \frac{4x^2 + 3y^2 + x^3y^3}{x^2 + y^2 + x^4y^4} &= \lim_{r \to 0}\frac{4 \left[ r\cos{(\theta)} \right] ^2 + 3 \left[ r\sin{(\theta)} \right] ^2 + \left[ r\cos{(\theta)} \right] ^3 \left[ r\sin{(\theta)} \right] ^3}{ r^2 + \left[ r\cos{(\theta)} \right] ^4 \left[ r\sin{(\theta)} \right] ^4 } \\ &= \lim_{r \to 0} \frac{ 4r^2 \cos^2{(\theta)} + 3r^2\sin^2{(\theta)} + r^6 \cos^3{(\theta)} \sin^3{(\theta)} }{ r^2 + r^8\cos^4{(\theta)}\sin^4{(\theta)} } \\ &= \lim_{r \to 0} \frac{4\cos^2{(\theta)} + 3\sin^2{(\theta)} + r^4\cos^3{(\theta)}\sin^3{(\theta)}}{1 + r^6\cos^4{(\theta)}\sin^4{(\theta)}} \\ &= \lim_{r \to 0} \frac{ \cos^2{(\theta)} + 3 + r^4\cos^3{(\theta)}\sin^3{(\theta)} }{1 + r^6\cos^4{(\theta)}\sin^4{(\theta)}} \\ &= \frac{\cos^2{(\theta)} + 3 + 0}{1 + 0} \\ &= \cos^2{(\theta)} + 3 \end{align*}

Since this value changes depending on the value of \displaystyle \begin{align*} \theta \end{align*}, the limit will be different depending on which path you take. Thus the limit does not exist.