# 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.

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

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

Therefore:

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

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

• Of course, this is not rigorous by any means, but the idea is there. – 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 $$\lim_{x \to 1^-} \frac{1}{(x-1)^{\frac{1}{3}}}=-\infty$$

• 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? – user532874 Aug 14 at 3:05
• $x\to 1~$ 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 $~0~$. – nmasanta Aug 14 at 3:19