Question

Please give explaination for this question ​

Answer 1

Step-by-step explanation:

Given:

$lim_{x->∞}\:\bigg( \frac{ {x}^{2} + 1 }{x + 1} - ax - b\bigg) = 0$

To find: Value of a and b

Solution:

Step 1: Take LCM

$lim_{x->∞}\: \bigg( \frac{ {x}^{2} + 1 - (ax + b)(x + 1) }{x + 1} \bigg) = 0$

Step 2: Simplify and separate coefficient of x²,x and constant term

$lim_{x->∞}\:\bigg( \frac{ {x}^{2} + 1 - a {x}^{2} - ax - bx - b }{x + 1} \bigg) = 0 \\ \\ lim_{x->∞}\: \bigg( \frac{ {x}^{2} (1 - a) - x(a + b) + 1 - b }{x + 1} \bigg) = 0$

Step 3: Put coefficient of x² to 0

Because when limit tends to infinity,it has finite only when numerator has the same or less degree as of denominator.

But in this case numerator has degree 2 but denominator has only degree 1.

Thus

$1 - a = 0 \\ \\ \boxed{a = 1}$

Step 4: Apply limit

$lim_{x->∞}\:: \bigg( \frac{ - x(a + b) + (1 - b) }{x + 1} \bigg) = 0$

take x common

$lim_{x->∞}\: \bigg( \frac{ - (a + b) + \frac{ (1 - b ) }{x}}{1 + \frac{1}{x}} \bigg) = 0$

$\: \bigg( \frac{ - (a + b) + 0}{1 + 0} \bigg) = 0$

$-(a+b)=0$

Step 5: Find value of b

$- (a + b) = 0 \\ \\ a + b = 0 \\ \\ put \: value \: of \: a \\ 1 + b = 0 \\ \\ \boxed{b = - 1}$

Final Answer:

a= 1 and b= -1

Hope it helps you.

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