How do you find the vertical asymptotes of this function?

\(f(x)=\dfrac{x-1}{x^{\frac23}-1}\)

I set the denominator equal to zero and get x = 1 or x = -1

But how do I know that x = 1 is not an asymptote but x = -1 is?

The question I really struggle with is how to find this:

\(\lim\limits_{ x \to 1^- }\dfrac{x-1}{x^{\frac23}-1}\qquad\text{and}\qquad\lim\limits_{ x \to 1^+ }\dfrac{x-1}{x^{\frac23}-1}\)

I haven't even gotten to the point of trying to find the limits as x approaches -1 from the positive and negative side. Also, I need to be able to find these limits without a graph.

hectictar Sep 10, 2018

#1**+3 **

When x is negative x^{2/3} =1 will have three possible solutions, two of them complex. If we are to stick with the reals, then we must do the squaring before the cube rooting. i.e. we must express x^{2/3} as (x^{2})^{1/3}.

When x approaches -1 from above then (x^{2})^{1/3 }-1 is negative, x - 1 is negative, so f(x) is positive, and approaches the limit +infinity.

When x approaches -1 from below then (x^{2})^{1/3 }-1 is positive, x - 1 is negative, so f(x) is negative, and approaches the limit -infinity.

For x = +1, use l'Hopital's rule to show the limit is 3/2.

Alan Sep 10, 2018