## anonymous 4 years ago Compute A= x^2+x/x+1 as x approaches +infinity and B= x^2+x/x^2+1 as x approaches -infinity

1. anonymous

By using l'hopital's rule, take the derivative of $x ^{2}+x = 2x+1$ and the derivative of $x+1 = 1$ $A = +\infty$ Similary for B = 1

2. anonymous

im not familiar with the l'hopital rule

3. anonymous

is there a way i can do it geometrically

4. anonymous

I don't think so. L'hopital's rule states that if you plug in the infinity into that equation and when the equations gives $\infty/\infty$ or 0/0 you take the derivative of the equations separetly.

5. anonymous

since im not familiar with this rule, can you walk me through it

6. anonymous

Did you read my post above? I gave you a little info about the rule. If you need detailed help just tell me.

7. anonymous

yes please, i need detailed help

8. anonymous

For example take the equation above $A = x ^{2}+x/x+1$ If you replace +infinity in the all of the x-s you get infinity/infinity which is an indeterminate form i.e. you cannot compute the answer. Because it is an indeterminate form you have to take the derivative of those equation seperatly, like i did in my first post.

9. anonymous

ok so what happens to the 2x+1/1

10. anonymous

replace infinity in that x which gives$2(\infty)+1/1 = \infty$

11. anonymous

oh ok i see

12. anonymous

so B= 2x+1/2x+1

13. anonymous

2(-inf)+1/2(-inf)+1 = -inf?

14. anonymous

No. The derivative of 1 = 0. The derivative of any constant is equal to 0

15. anonymous

i understand now thanks