+0  
 
+1
162
3
avatar+7336 

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
edited by hectictar  Sep 10, 2018
 #1
avatar+27128 
+2

When x is negative x2/3 =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 x2/3 as (x2)1/3.

 

When x approaches -1 from above then  (x2)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  (x2)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
edited by Alan  Sep 10, 2018
 #2
avatar
+2

You can calculate this limit without l'hopital.

 

Hint: a3-1=(a-1)*(a2+a+1)

 

a2-1=(a-1)*(a+1)

Guest Sep 10, 2018
 #3
avatar+7336 
+1

Thank you, I finally get it now  laugh

hectictar  Sep 10, 2018

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