How come this limit is equal 2

How come

LiM of X that goes to plus INFINITE of ( 2x + 1 ) OVER ( x - 2 ) is equal 2???????

How's that possible? Shouldn't it be indeterminate?

(2x-1)/(x-2)=
2(x-2)/(x-2)+3/(x-2)=
2+3/(x-2)

2+3/(x-2) -> 2 when x->infnty

lim x → ∞ [(2x + 1) / (x - 2)]

lim x → ∞ [(2x/x + 1/x) / (x/x - 2/x)]

lim x → ∞ [(2 + 1/x) / (1 - 2/x)]

plug in infinity and as x gets really large 1/x and 2/x approaches 0.

= [(2) / 1]

= 2

lim x->infi (2x+1)/(x-2) dividedby x , both deno.and nume.

lim 1/x->0 (2+1/x)/(1-2/x)= (2+0)/(1-0)=2 ,x->inf then 1/x->0

answer is:
( 2x + 1 ) / ( x - 2 ) = ( 2 + 1/x ) / ( 1 - 2/x )

Lim ( 2 + 1/x ) / ( 1 - 2/x ) = 2/1
x→∞

The limit is 2 as you can't divide by 0. (2-2=0) google inverse function on images.

When x → ∞,

(2x+1) → 2x,

&

(x-2) → x;

So,

[(2x+1)/(x-2)] → (2x/x) → 2;

( 2x + 1 ) / ( x - 2 ) = ( 2 + 1/x ) / ( 1 - 2/x )

Lim ( 2 + 1/x ) / ( 1 - 2/x ) = 2/1
x→∞