Find all the different limits in + infinity

[JPC]

Homework Statement

f(x) = sqroot(ax^2 + 1 ) - x

find all the different limits in + infinity that f can have with all the different "a" values

Homework Equations

The Attempt at a Solution

i don't know if i did something wrong here :

- if a = 0 then f = 1 - x, limit : - infinty

- if a < 0 then no f value in + infinite
, (except if a = ' - 1 / +infnte', limit : - infinte)

- if a > 0 then

lim infinite ( f(x) ) = lim +infinite ( root (ax^2 . (1 + 1 / ax^2) ) - x )
= lim +infinte ( root(a) . x - x)
= lim +infinte ( (root(a) - 1) x)

so if a = 1, limit is 0
if a > 1 , limit is + infinity
if 0 < a < 1, limit is - infinity


PS ; if badly wrote, ect, I am sorry , but i was on a pc with a weird keyboard that really thrustrated me

 

[EnumaElish]

Can you explain "if a < 0 then no f value in + infinite"?

How do you go from "lim +infinite ( root (ax^2 . (1 + 1 / ax^2) ) - x )" to "lim +infinte ( root(a) . x - x)"?

 

[JPC]

hum

- considering the equation g(x) = ax² + 1
if a < 0, the, when x reaches +infinite, g(x) reaches -infinite
and since we only want to study f(x) in R , and not in complex numbers, we don't consider to be a value of f(x) when x reaches +infinite

(im not sure if you say 'reach' in english , but in french we call it 'tend vers')

- now if a > 0 :

lim + infinite ( f(x) )
= lim +infinite ( sqroot (ax^2 . (1 + 1 / ax^2) ) - x )

and

lim + infinite (1 + 1 / ax^2) = 1 + 0 = 1

so

lim +infinite ( sqroot (ax^2 . (1 + 1 / ax^2) ) - x )
= lim +infinite ( sqroot (ax^2 ) - x )
= lim +infinite ( sqroot(a)* x - x )
= lim +infinite ( (sqroot(a) -1 ) * x )

so if
(sqroot(a) -1 ) > 0
then
lim + infinite ( f(x) ) = +infinite

and if
(sqroot(a) -1 ) < 0
then
lim + infinite ( f(x) ) = -infinite

finding (sqroot(a) -1 ) = 0
sqroot(a) = 1
a = 1

so if
a>1
then
(sqroot(a) -1 ) > 0
lim + infinite ( f(x) ) = +infinite

and if
0<a<1
then
(sqroot(a) -1 ) < 0
lim + infinite ( f(x) ) = -infinite

 

[JPC]

and also :

its not in my program yet (will be in a few months) but is this true for a < 0 : ?

since
lim +oo(f(x)) = lim +infinite ( (sqroot(a) -1 ) * x )

a = -b , a<0

sqroot(-b) - 1 = i * sqroot(b) - 1

so lim +infinite ( (i* sqroot(b) -1 ) * x )

so lim +oo(f(x)) = (h, +oo)

- Now let's find that angle h :

r² = (-1)² + (root(b))² = b+1
r = root(b+1)
h = z + pi/2

by looking at the rectangular triangle of sides : 1, root(b), root(b+1) :

sin(z) = 1 / root(b+1)

z = sin^-1(1 / root(b+1))

so h = pi/2 + sin^-1(1 / root(b+1))

So when x-> +oo, and a<0 :
lim lim+oo (f(x) ) = (pi/2 + sin^-1(1 / root(b+1)) , +oo)


ok, i haven't worked at all with complex numbers atm, i just found that with the basic knowledge and understanding i have of complex numbers ( i = root(-1), i² = - 1, and the graphical represantation of complex numbers)

So , if i did anything wrong here, could you tell me please

 

[EnumaElish]

im not sure if you say 'reach' in english , but in french we call it 'tend vers'

Approaches, tends toward (formal) or goes to (informal).

 

[JPC]

yes, but did i do any mistake in what i did ?

i mean did i use a wrong method , and just got lucky to get the right answer ? (i checked the limits with my calculator)
Because , what counts more in french schools is the way you find the answer rather than the answer itself