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

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.

  • 2
    $\begingroup$ I believe it just diverges to negative infinity. $\endgroup$ – JG123 Aug 14 at 2:50
  • $\begingroup$ @JG123 How did you arrive at that conclusion? Could you explain it in an answer block for this question? $\endgroup$ – user532874 Aug 14 at 2:54
  • $\begingroup$ You can find the limit by substituting x=0.992,0.999,0.999999 and you’ll see that $\endgroup$ – Isaac YIU Math Studio Aug 14 at 2:56
  • $\begingroup$ The denominator approaches $0$ from the negative side, while the numerator remains constant. Therefore the fraction approaches negative infinity. $\endgroup$ – abiessu Aug 14 at 2:57

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:

$$\lim_{u\to -\infty}\sqrt[3]{u}=-\infty.$$


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


$\lim_{x\to 1^-}$ $\frac{1}{(x-1)^{1/3}}$=$\frac{1}{\epsilon}$, 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.

  • $\begingroup$ Of course, this is not rigorous by any means, but the idea is there. $\endgroup$ – JG123 Aug 14 at 3:09

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

  • $\begingroup$ The only part I don't get is how do we know the denominator is approaching 0? Is this because substituting 1 into the denominator yields 0 for the denominator? $\endgroup$ – user532874 Aug 14 at 3:05
  • $\begingroup$ $x\to 1~$ does not mean that you have to substitute $~1~$ in $~x~$. Actually by $~x \to 1~$ , you have to consider the value of $~x~$ which is nearly equal to $~1~$(difference between $~x-1~$ is nearly equal to $~0~$ ). And for this case the denominator is so small that you can consider it is nearly equal to $~0~$. $\endgroup$ – nmasanta Aug 14 at 3:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.