Question

$li{m}_{x\to \infty }x(\arctan \frac{x+1}{x+2}-\arctan \frac{x}{x+2})$ is equal to

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $1$
• D. D.N.E

Answer 1

Answer: (A)

$\frac{1}{2}$

${lim}_{x\to \infty }x(\arctan \left(\frac{x+1}{x+2}\right)-\arctan \left(\frac{x}{x+2}\right))={lim}_{x\to \infty }x\left(\arctan \left(\frac{\frac{x+1}{x+2}-\frac{x}{x+2}}{1+\frac{x(x+1)}{{(x+2)}^2}}\right)\right)={lim}_{x\to \infty }x\left(\arctan \left(\frac{(x+2)}{2x^2+5x+4}\right)\right)={lim}_{x\to \infty }\frac{\arctan \left(\frac{x+2}{2x^2+5x+4}\right)}{\left(\frac{1}{x}\right)}$

$={lim}_{x\to \infty }\frac{\frac{1}{(1+\frac{{(x+2)}^2}{{(2x^2+5x+4)}^2})}\cdot \frac{(2x^2+5x+4)-(x+2)(4x+5)}{{(2x^2+5x+4)}^2}}{(-\frac{1}{x^2})}$ [By L Hospital's Rule]

$={lim}_{x\to \infty }\frac{(2x^2+5x+4-4x^2-13x-10)}{{(2x^2+5x+4)}^2+{(x+2)}^2}(-x^2)={lim}_{x\to \infty }\frac{2x^4+8x^3+6x^2}{{(2x^2+5x+4)}^2+{(x+2)}^2}={lim}_{x\to \infty }\frac{2+\frac{8}{x}+\frac{6}{x^2}}{{(2+\frac{5}{x}+\frac{4}{x^2})}^2+{(\frac{1}{x}+\frac{2}{x^2})}^2}=\frac{2+0+0}{{(2+0+0)}^2+(0+0)}=\frac{1}{2}$