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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.

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    $\begingroup$ 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
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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:

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

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

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  • $\begingroup$ Of course, this is not rigorous by any means, but the idea is there. $\endgroup$ – JG123 Aug 14 at 3:09
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$~x\to 1^-~$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 $$\lim_{x \to 1^-} \frac{1}{(x-1)^{\frac{1}{3}}}=-\infty$$

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  • $\begingroup$ 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

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