We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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**+2 **

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