Question

Help me please to find the limit of the function attached; x approaches to -inf. Help me please to find the limit of the function attached; x approaches to -inf. @Mathematics

Answer 1

hello joe!

Answer 2

hey satellite :)

Answer 3

I know the answer is -inf

Answer 4

I dont know how to get that!

Answer 5

i can't really read it, because i am not sure what that first x is doing there, but gimmick it to multiply by the conjugate in the form of
\[\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\] and then take the limit

Answer 6

x is multiplying what is inside the parenthesis satellite. Hope this new attachment helps you.

Answer 7

look..

you can observe when x is very very big the 1 is irrelevant so sqrt(x^2 +1) = x' , but less then origianl x...

so we have:

take the limit x*x' -x^2

thats result infinite because x^2 grows up faster then x*x'

Answer 8

dont get it. can you put it in other words, please

Answer 9

I applied the conjugate but it didint work.

Answer 10

lets take some numbers and put then in y=sqrt(x^2 +1).

x=10 y = 10,04987
x=100 y= 100,004999
x= 1000 y = 1000,0005
.
.
.
x=100 000 000 y = 100 000 000

until here its clear?

Answer 11

oh! the comma is decimal separator...

Answer 12

can I do this?

I know if I apply the limit to the expression under the root, it is equal to infinite.

I apply the limit to the rest of the expression so it looks like this.

=-inf(inf - (-inf))
= -inf(inf + inf)
=-inf(inf) coz inf + inf is equal to inf, right?
=-inf.

This is the only way I get the correct answer.

Answer 13

remember that x approaches -inf

Answer 14

mathematically speaking you couldnt make operations with infinity... its not defined! (remember L'hopital!)

Answer 15

i dont know other way to explain.. sorry..=/

Answer 16

this is correct, what you said
can I do this? I know if I apply the limit to the expression under the root, it is equal to infinite. I apply the limit to the rest of the expression so it looks like this. =-inf(inf - (-inf)) = -inf(inf + inf) =-inf(inf) coz inf + inf is equal to inf, right? =-inf.
it is not "indeterminate form"

Answer 17

then according to satellite

+inf-inf =zero ! oO

Answer 18

but you have (inf -(-inf))= inf + inf = inf, right, or definitively not?

Answer 19

no no you do not have that
you have
\[-\infty\times (\infty +\infty)\]

Answer 20

oh right, the second thing you said is correct

Answer 21

\[\infty -\infty\] you don't know, but
\[\infty+\infty\] you do

Answer 22

inf + inf = inf

Answer 23

yes that is right

Answer 24

i hope i did not say
\[-\infty + \infty=0\] because that is way wrong. but
\[-\infty\times \infty =-\infty\]

Answer 25

so you mean that the way I worked out the problem before is right? or partially right?