Find the constants a and b from the condition : (i) lim_(x→∞) (((x^2 +1)/(x+1))−ax−b)=0 (ii) lim


Question and Answers Forum

All Questions Topic List
Limits Questions
Previous in All Question Next in All Question
Previous in Limits Next in Limits
Question Number 122725 by liberty last updated on 19/Nov/20

$F i n d t h e c o n s t a n t s a a n d b f r o m t h e$ $c o n d i t i o n :$ $(i) \lim_{x \to \infty} (\frac{x^{2} + 1}{x + 1} - a x - b) = 0$ $(i i) \lim_{x \to \infty} (\sqrt{x^{2} - x + 1} - a x - b) = 0$
Commented by Dwaipayan Shikari last updated on 19/Nov/20
$lim_(x→∞) (√(x^2 −x+1))−ax−b lim_(x→∞) x((√(1−(1/x)+(1/x^2 )))−a−(b/x)) =x(1−(1/(2x))+(1/(2x^2 ))−a−(b/x))=0 If 1−a=0⇒a=1 and then it becomes −(1/2)+(1/(2x))−b=0⇒ x→∞ b=−(1/2) so { ((a=1)),((b=−(1/2))) :}$
$\lim_{x \to \infty} \sqrt{x^{2} - x + 1} - a x - b$ $\lim_{x \to \infty} x (\sqrt{1 - \frac{1}{x} + \frac{1}{x^{2}}} - a - \frac{b}{x})$ $= x (1 - \frac{1}{2 x} + \frac{1}{2 x^{2}} - a - \frac{b}{x}) = 0$ $I f 1 - a = 0 \Rightarrow a = 1$ $a n d t h e n i t b e c o m e s - \frac{1}{2} + \frac{1}{2 x} - b = 0 \Rightarrow$ $x \to \infty b = - \frac{1}{2} s o {\begin{cases} a = 1 \\ b = - \frac{1}{2} \end{cases}$
Commented by bemath last updated on 19/Nov/20
$(i) lim_(x→∞) (((x^2 +1)/(x+1))−(((ax+b)(x+1))/(x+1))) = lim_(x→∞) (((x^2 +1−(ax^2 +(a+b)x+b))/(x+1))) = lim_(x→∞) ((((1−a)x^2 −(a+b)x+1−b)/(x+1))) because the limit equal to 0 , it meant { ((1−a = 0⇒a=1)),((−(a+b)=0⇒b=−1)) :}$
$(i) \lim_{x \to \infty} (\frac{x^{2} + 1}{x + 1} - \frac{(a x + b) (x + 1)}{x + 1}) =$ $\lim_{x \to \infty} (\frac{x^{2} + 1 - ({a x}^{2} + (a + b) x + b)}{x + 1}) =$ $\lim_{x \to \infty} (\frac{(1 - a) x^{2} - (a + b) x + 1 - b}{x + 1})$ $b e c a u s e t h e l i m i t e q u a l t o 0, i t m e a n t$ ${\begin{cases} 1 - a = 0 \Rightarrow a = 1 \\ - (a + b) = 0 \Rightarrow b = - 1 \end{cases}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum