Confusion over a limit. Different ways of solving give different answers? Qn: If it is given that
$$
\lim_{x\to\infty} \frac{x^2 - x - 2}{x + 1} - ax - b = 1
$$
then a and b must be?
Now, I tried doing this by 2 methods.
Method 1:
$$ \frac{x^2 - x - 2}{x + 1} - ax - b $$
$$ = (x - 2) - ax - b $$
Since the limit is finite, $a$ must be $= 1$ and so, $b = -3$
Method 2:
$$ \frac{x^2 - x - 2}{x + 1} - ax - b $$
$$ = \frac{x^2(1 - \frac 1 x - \frac2 {x^2})}{x(1 + \frac1x)} - ax - b $$
$$ = \frac{x - 1 - \frac2x}{(1 + \frac1x)} - ax - b $$
as $x \to \infty$, we have the above expression
$$ = x - ax - b$$
So, $a = 1$ and $b = -1$
Which of the above is correct?
 A: Method $1$ is correct
In Method $2$, you cannot apply limit to a part of the fraction. So
$$ = \frac{x^2(1 - \frac 1 x - \frac2 {x^2})}{x(1 + \frac1x)} - ax - b $$
is not equal to 
$$ = x - ax - b$$
In the above step you have applied limit to just a part of that fraction leaving behind $\dfrac{x^2}{x}$.
hope the answer is clear !
A: In method 2 you are losing too much information. In particular, you lose the linear and constant term in the numerator of the fraction.
To be more particular,
you need to write
$\frac{x^2 - x - 2}{x + 1} - ax - b
=\frac{x^2 - x - 2 - (x+1)(ax+b)}{x + 1}
=\frac{x^2 - x - 2 - (ax^2+x(a+b)+b)}{x + 1}
=\frac{x^2(1-a) - x(1+a+b) - 2 -b}{x + 1}
$.
If you want the limit of this to exist,
you must have $a = 1$, or else the $x^2$ term
will cause the fraction to be unbounded.
The fraction then becomes
$\frac{- x(2+b) - 2 -b}{x + 1}
$.
Any value of $b$ allows the limit to exist,
since it is of the form
$\frac{linear}{linear}$.
To have the limit be $1$,
we will ignore the fortuitous factorization
that can be done with the numerator
and just look at the coefficients of $x$.
For the limit to be $1$,
the coefficients of $x$ is the numerator and denominator must be equal,
so that $1 = -(2+b)$,
or $b = -3$.
Note that the constant terms in this fraction do not matter.
In other words,
if $c$ and $d$ are any real numbers,
to make $\lim_{x \to \infty} \frac{- x(2+b)+c}{x + d} = 1$,
you must choose $b = -3$.
The limit then becomes
$\lim_{x \to \infty} \frac{x+c}{x + d}$,
and this limit is $1$ for any $c$ and $d$.
To see this,
note that
$\frac{x+c}{x + d}-1
= \frac{x+c-(x+d)}{x + d}
= \frac{c-d}{x + d}
$
and this goes to $0$
as $x \to \infty$.
