Given-lim-x-x-2-x-3-x-b-4-find-the-value-of-b-2-1-

Question Number 120720 by bramlexs22 last updated on 02/Nov/20

Given \lim_{x \to \infty} \sqrt{x + 2 \sqrt{x} + 3} - \sqrt{x} + b = 4

find the value of b^{2} + 1

Answered by Dwaipayan Shikari last updated on 02/Nov/20

\sqrt{x} (\sqrt{1 + \frac{2}{\sqrt{x}} + \frac{3}{x}} - 1 + \frac{b}{\sqrt{x}}) = 4

\lim_{x \to \infty} \sqrt{x} (1 + \frac{1}{\sqrt{x}} + \frac{3}{2 x} - 1 + \frac{b}{\sqrt{x}}) = 4

\sqrt{x} (\frac{1 + b}{\sqrt{x}}) = 4

b = 3

b^{2} + 1 = 10

Answered by john santu last updated on 02/Nov/20

\lim_{x \to \infty} \sqrt{x + 2 \sqrt{x} + 3} - (\sqrt{x} - b) = 4

\lim_{x \to \infty} \sqrt{x + 2 \sqrt{x} + 3} - \sqrt{{(\sqrt{x} - b)}^{2}} = 4

\lim_{x \to \infty} \sqrt{x + 2 \sqrt{x} + 3} - \sqrt{x - 2 b \sqrt{x} + b^{2}} = 4

\frac{2 - (- 2 b)}{2.1} = 4 \Rightarrow 2 + 2 b = 8 \Rightarrow b = 3 \land b^{2} + 1 = 10