Question 1029969
<pre><b>
{{{f(x)= (2x^2-7)/(x+2^"")}}}

   <u>     2x-4</u>
x+2)2x²+0x-7
    <u>2x²+4x</u>
       -4x-7
       <u>-4x-8</u>
           1

So {{{f(x) = 2x-4+1/(x+2)}}}

Our claim is that the slant asymptote is 

g(x)=2x-4.

So we prove that the difference between 
f(x) and g(x)
as x approaches ħinfinity is 0.

{{{matrix(1,8,

matrix(2,1,lim,"x->oo"),

(f(x)^""-g(x)),

""="",
matrix(2,1,lim,"x->oo"),

((2x^2-7)/(x^""+2)-(2x-4)),

""="",

matrix(2,1,lim,"x->oo"),
(
(2x^2-7)/(x^""+2)-(2x^""-4)*(x^""+2)/(x^""+2)
)

)}}}{{{""=""}}}

{{{matrix(1,8,

matrix(2,1,lim,"x->oo"),

(
(2x^2-7)/(x^""+2)-(2x^2-8)/(x^""-2)
),

""="",

matrix(2,1,lim,"x->oo"),

(
(2x^2-7-(2x^2-8))/(x^""-2)
),

""="",

matrix(2,1,lim,"x->oo"),

((2x^2-7-2x^2+8)/(x^""-2))

)}}}{{{""=""}}}

{{{matrix(1,4,
matrix(2,1,lim,"x->oo"),

(1/(x^""-2)),
""="",
0)}}}


Similar for {{{matrix(1,4,

matrix(2,1,lim,"x->-oo"),

(f(x)^""-g(x)),

""="", 0)}}}

Edwin</pre>