Question

find all values of c where the limit exists\lim _{x\to \infty }\left(\frac{x^3-1}{x^c+1}\right)

Answer 1

I can't read the equation well but I'm pretty sure that
Desmos.com solves some kind of equations and GeoGebra.com solves pretty much all kinds

Answer 2

@Will.H the limit as x--> infinity of (x^3-1) / (x^c+1)

Answer 3

The relevant fact is that −|f(x)|≤f(x)≤|f(x)| −|f(x)|≤f(x)≤|f(x)| for any function f f. Once you see how to apply that here, the problem becomes easy to solve.

Answer 4

@myininaya pm me once you done here

Answer 5

\[\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)}\]
Let f(x) and g(x) be polynomials.
If f(x) and g(x) share the same degree, then the limit exists.
If f(x) has a higher degree than g(x), then the limit does not exist.
If f(x) has a lower degree than g(x), then the limit exists.

This though does not take into account for negative values of c; you know when g(x) is not a polynomial. So you are left to consider those cases now.
Try assuming c=-p where p>0 and see what happens...

Answer 6

https://www.mathsisfun.com/calculus/limits-infinity.html
Here is a good reference to what I have mentioned above.

Answer 7

find the limit when
1. c=3
2.c>3
3. c<3