Problem formulation:
Determine constants a_i and b_i ( i = 1,2 ) from the following conditions:
x→−∞lim(x2−x+1−a1x−b1)=0x→+∞lim(x2−x+1−a1x−b1)=0
No particularly hard to grasp insights are required to ''work'' with the numbers, which represents the first part of the solution, though it is quite lengthy, so I am simply going to post the picture of the solution below.
The part I have trouble with is the very end: actually finding the solutions from the simplified equation(s).
The last line of the simplified equation has the following dividend implies that the simplified dividend has to be 0, so:
x→−∞lim(1−a12)x−(1+2a1b1)+(1−b12)x1=0
From this, the authors deduce ( as shown in the official solution picture ) the following equations:
1−a12=0−1+a1−(1+2a1b1)=0
...from which they deduce that a_1 = -1 and b_1 = 1\2
First, I don't understand the second of the 2 equations; where did they get that divisor ( -1 + a_1 ) from ??
Second, doesn't 1 - a_1^2 = 0 imply that a_1 = +- 1, and not only -1 ( since to my knowledge we have no domain restrictions for a_1 anywhere in the problem ).
From my understanding, the second equation should just be 1 + 2a_1b_1 = 0, from which b_1 = -1 \ 2a_1, thus b_1 = +- 1\2. Why do they opt for only one of the solutions to a_1 and b_1 and not the other ?
The other confusing part is their solution for a_2 and b_2. They say the solution is analogous for a_2 and b_2, and that it's thus apparent that a_2 = 1 and b_2 = -1/2.
I'm aware that this definitely has to do with the different limits of the 2 other than that ( and the given unknowns ) identical equations, but I can't quite figure out how positive and negative limits affect any of this...
The official solution gives just one turn to the limit ( upon resolution of abs(x), which ultimately ( to my mind ) ends up being irrelevant anyway ). So, why does the positive and negative infinite limit of x determine whether we should pick positive or negative ''version'' of a for a_1 and a_2, and the corresponding ''version'' of b ?
I would appreciate all help with this.
EDIT: I now understand the solution: the expression isn't defined for -1 + a_1 = 0, so a_1 cannot equal 1; the absolute value of x's relation to the limit now makes sense, since the divisor of the expression will be 1 + a_1.
Determine constants a_i and b_i ( i = 1,2 ) from the following conditions:
x→−∞lim(x2−x+1−a1x−b1)=0x→+∞lim(x2−x+1−a1x−b1)=0
No particularly hard to grasp insights are required to ''work'' with the numbers, which represents the first part of the solution, though it is quite lengthy, so I am simply going to post the picture of the solution below.
The part I have trouble with is the very end: actually finding the solutions from the simplified equation(s).
The last line of the simplified equation has the following dividend implies that the simplified dividend has to be 0, so:
x→−∞lim(1−a12)x−(1+2a1b1)+(1−b12)x1=0
From this, the authors deduce ( as shown in the official solution picture ) the following equations:
1−a12=0−1+a1−(1+2a1b1)=0
...from which they deduce that a_1 = -1 and b_1 = 1\2
First, I don't understand the second of the 2 equations; where did they get that divisor ( -1 + a_1 ) from ??
Second, doesn't 1 - a_1^2 = 0 imply that a_1 = +- 1, and not only -1 ( since to my knowledge we have no domain restrictions for a_1 anywhere in the problem ).
From my understanding, the second equation should just be 1 + 2a_1b_1 = 0, from which b_1 = -1 \ 2a_1, thus b_1 = +- 1\2. Why do they opt for only one of the solutions to a_1 and b_1 and not the other ?
The other confusing part is their solution for a_2 and b_2. They say the solution is analogous for a_2 and b_2, and that it's thus apparent that a_2 = 1 and b_2 = -1/2.
I'm aware that this definitely has to do with the different limits of the 2 other than that ( and the given unknowns ) identical equations, but I can't quite figure out how positive and negative limits affect any of this...
The official solution gives just one turn to the limit ( upon resolution of abs(x), which ultimately ( to my mind ) ends up being irrelevant anyway ). So, why does the positive and negative infinite limit of x determine whether we should pick positive or negative ''version'' of a for a_1 and a_2, and the corresponding ''version'' of b ?
I would appreciate all help with this.
EDIT: I now understand the solution: the expression isn't defined for -1 + a_1 = 0, so a_1 cannot equal 1; the absolute value of x's relation to the limit now makes sense, since the divisor of the expression will be 1 + a_1.