Limits of Functions and Asymptotes

  • Thread starter Thread starter Econometricia
  • Start date Start date
  • Tags Tags
    Functions Limits
Econometricia
Messages
33
Reaction score
0
1. I am concerned with finding the discontinuities of a functin say x3+3x2+2x / (x-x3)
2. I am having issues with classifying for the type of discontinuities. Finding them is not an issue.
3. Also when finding horizontal asymptotes F(x) = 4 / (2e-x +1) I understnad that the HA are found by taking the limit of the function as X--->-INF/INF , but why is one of the HA 4?

Thank You =).

Homework Statement


Homework Equations


The Attempt at a Solution

 
Physics news on Phys.org
Econometricia said:
1. I am concerned with finding the discontinuities of a functin say x3+3x2+2x / (x-x3)



2. I am having issues with classifying for the type of discontinuities. Finding them is not an issue.
Do you mean classifying them into vertical or horizontal asymptotes? The vertical asymptotes are generally the numbers that make the denominator zero, that don't also make the numerator zero. To find them, factor both the numerator and denominator. The numbers that make the denominator zero are x = -1, x = 0, and x = 1. x = 0 is NOT a vertical asymptote, because the numerator is also zero when x = 0.
Econometricia said:
3. Also when finding horizontal asymptotes F(x) = 4 / (2e-x +1) I understnad that the HA are found by taking the limit of the function as X--->-INF/INF , but why is one of the HA 4?
As x --> infinity, e-x --> 0, so the denominator --> 1, and the overall fraction --> 4.
Econometricia said:
Thank You =).
 
Thank You for your help. I was actually meaning the discontinuities as Jump,Removable, Infinite. So far I have understood that only piece wise functions can have a Jump. Infinite limits are when the lim of F(x) as X-->A = INF/-INF. And removable is when the limit exists , but is not defined. Am I correct?
 
Econometricia said:
Thank You for your help. I was actually meaning the discontinuities as Jump,Removable, Infinite. So far I have understood that only piece wise functions can have a Jump. Infinite limits are when the lim of F(x) as X-->A = INF/-INF. And removable is when the limit exists , but is not defined. Am I correct?
Close. A removable discontinuity occurs when lim_{x \to a} f(x) exists (that means both one-sided limits exist), but f is not defined at a.

In your first example, I believe that there is a removable discontinuity at x = 0.
 
Yea, that is correct. =) Thank you sir.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top