Question

Compute A= x^2+x/x+1 as x approaches +infinity and B= x^2+x/x^2+1 as x approaches -infinity

Answer 1

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

Answer 2

im not familiar with the l'hopital rule

Answer 3

is there a way i can do it geometrically

Answer 4

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.

Answer 5

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

Answer 6

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

Answer 7

yes please, i need detailed help

Answer 8

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.

Answer 9

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

Answer 10

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

Answer 11

oh ok i see

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

i understand now thanks