Showing that $\lim_{(x,y,z)\to(0,e,-2/3)}\frac{\sqrt[3]{x^4}\left(1-\ln y\right)^\frac 53\left(z+\frac23\right)}{x^4+(y-e)^4+(3z+2)^4}$ does not exist

Question

Show that the following limit does not exist: Solve using the method described below.

$$\lim_{(x,y,z)\rightarrow (0,e,-2/3)} \frac{\sqrt[3]{x^4}\left(1-\ln(y)\right)^\frac 53 \left(z+\frac23\right)}{x^4+(y-e)^4+(3z+2)^4}$$

What I've done so far...

In order to get everything in terms of 1 variable, I am testing all possible straight lines through the point $(0,e,-2/3)$ by creating the parametric equations of $x$, $y$, and $z$ in terms of $t$. The equations I've used are as follows:

$$x=at$$ $$y = e+bt$$ $$z = \frac{-2}{3}+ct$$

where a, b, and c are constants that change and represent all possible lines passing through the given point. After substituting these new equations I am given:

$$\lim_{(t)\rightarrow (0)} \frac{\sqrt[3]{at}^4\left(1-\ln(e+bt)\right)^\frac53 \left(\frac{-2}{3}+ct+\frac23\right)}{(at)^4+(e+bt-e)^4+\left(3\left(\frac{-2}{3}+ct\right)+2\right)^4}$$

After some algebraic simplification, I am left with,

$$\lim_{(t)\rightarrow (0)} \frac{a^\frac43c(1-\ln(e+bt))^\frac53}{t^\frac53(a^4+b^4+3^4c^4)}$$

This is where I am stuck because I cannot find a way to simplify this final expression. L'Hopital's rule doesn't seem to work because you are continually left in indeterminate form and I can't think of a way to work around the natural log to a power. After plugging the above equation into Wolfram, it says that this limit is equivalent to

$$\frac{a^\frac43(-b^\frac53)c}{e^\frac53\left(a^4+b^4+3^4c^4\right)}$$

as t approaches 0. This answer makes sense, because approaching along different lines with different values of a, b, and c through the point give you different limits, proving that the original limit to the original equation does not exist.

If someone could explain how the final answer is developed from the limit, that would answer my question. Additionally, if any of my algebra or simplification was wrong along the way, please let me know. Finally, this is the method that we were instructed to use when proving that the limit does not exist.

Answer 1

Let $f(x,y,z):=\frac{\sqrt[3]{x^4}(1-ln(y))^\frac 53 (z+2/3)}{x^4+(y-e)^4+(3z+2)^4}.$

For $x>0$ we have

$f(x,e,- \frac{2}{3})=\frac{1}{x^{8/3}} \to \infty$ as $x \to 0^+.$

Answer 2

Use Taylor series around $t=0$ $$\log(e +b t)=1+\frac{b t}{e}-\frac{b^2 t^2}{2 e^2}+O\left(t^3\right)$$ $$1-\log(e +b t)=-\frac{b t}{e}+\frac{b^2 t^2}{2 e^2}+O\left(t^3\right)$$ $$(1-\log (e+b t))^{5/3}=\frac{(-b)^{5/3} t^{5/3}}{e^{5/3}}+\frac{5 (-b)^{8/3} t^{8/3}}{6 e^{8/3}}+O\left(t^{11/3}\right)$$

Answer 3

I'd first do the substitutions $x=X^3$, $y=(1+Y)e$ and $z=Z-2/3$, so the limit becomes $$ \lim_{(X,Y,Z)\to(0,0,0)}\frac{-X^4(\ln(1+Y))^{5/3}Z}{X^{12}+e^4Y^4+81Z^4} $$ Now, for $X=at$, $Y=b^3t^3$, $Z=ct^3$, we get $$ \frac{-a^4ct^7(\ln(1+b^3t^3))^{5/3}}{t^{12}(a^{12}+e^4b^{12}+81c^4)} $$ The logarithmic term is $$ (\ln(1+b^3t^3))^{5/3}=(b^3t^3+o(t^3))^{5/3}=b^5t^5+o(t^5) $$ so we finally get $$ \frac{-a^4b^5c+o(1)}{a^{12}+e^4b^{12}+81c^4} $$