Solving $\lim_{x \to 1^-} \frac{1}{(x-1)^{\frac{1}{3}}}$ without a calculator Problem:

$$\lim_{x \to 1^-} \frac{1}{(x-1)^{\frac{1}{3}}}$$

Without a calculator, what is the simplest method to solving this limit? I multiplied $\frac{1}{(x-1)^{\frac{1}{3}}}$ by $\frac{(x-1)^{2/3}}{(x-1)^{2/3}}$ to get $\frac{(x-1)^{2/3}}{(x-1)}$. I then rewrote the limit as $\lim_{x \to 1^-} \frac{(x-1)^{2/3}}{(x-1)}$ so I could get an indeterminate form. I then tried using l'hospital's rule but I got an undefined answer.
 A: Let $t=x-1$. Then, $x\to 1^{-}\implies t\to 0^{-}$. So, we have the following:
$$
\lim_{x\to 1^{-}}\frac{1}{\left(x-1\right)^{\frac{1}{3}}}=
\lim_{t\to 0^{-}}\frac{1}{\sqrt[3]{t}}=
\lim_{t\to 0^{-}}\sqrt[3]{\frac{1}{t}}.
$$
The behavior of the function $f(t)=\frac{1}{t}$ is well known. As $t$ approaches $0$ from the left, the functional value is becoming an increasingly large negative number. So, again, let $u=\frac{1}{t}$. Then, $t\to 0^{-}\implies u\to -\infty$. Our limit now looks like this:
$$\lim_{u\to -\infty}\sqrt[3]{u}.$$
And what does the function $f(u)=\sqrt[3]{u}$ approach as its argument goes to negative infinity? Well, it also goes negative infinity:
$$\lim_{u\to -\infty}\sqrt[3]{u}=-\infty.$$
Therefore:
$$\lim_{x\to 1^{-}}\frac{1}{\left(x-1\right)^{\frac{1}{3}}}=-\infty.$$
A: $~x\to 1^-~$i.e., $~x~$ approaches $~1~$ from the left and  so $$~x\lt 1\implies x-1\lt 0\implies (x-1)^{\frac{1}{3}}\lt 0~$$
Therefore the denominator is a negative quantity  approaching $~0~$ from the left. 
Hence $$\lim_{x \to 1^-} \frac{1}{(x-1)^{\frac{1}{3}}}=-\infty$$
A: $\lim_{x\to 1^-}$ $\frac{1}{(x-1)^{1/3}}$=$\frac{1}{\epsilon}$, where $\epsilon$ is an infinitesimal number just less than $0$. This can be seen by plugging in a number just less than $1$ into $(x-1)^{1/3}$. Thus, we have that $\frac{1}{\epsilon}$=-$\infty$ because we are dividing 1 by a tiny negative number.
