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.

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:
$$\lim_{x\to 1^{-}}\frac{1}{\left(x-1\right)^{\frac{1}{3}}}=-\infty.$$
