calculus

Question

$\lim_{x \to \infty}\left (\frac{1}{1-x^{2}} + \frac{2}{1-x^{2}}+\dots+\frac{x}{1-x^{2}}\right )$ is equal to

(a) $0$

(b) $-1/2$

(c) $1/2$

(d) None of the above

Answer 1

$\lim_{x \to \infty}\left (\frac{1}{1-x^{2}} + \frac{2}{1-x^{2}}+.....+\frac{x}{1-x^{2}}\right )$

$\lim_{x \to \infty}\frac{1}{1-x^{2}} (1+2+3....+x)$

=$\lim_{x \to \infty}\frac{1}{1-x^{2}} \frac{x(x+1)}{2}$

=$\lim_{x \to \infty}\frac{1}{(1-x)(1+x)} \frac{x(x+1)}{2}$

=$\frac{1}{2} \times \lim_{x \to \infty}\frac{1}{(\frac{1}{x}-1)}$

put $x= \infty$

=$-\frac{1}{2}$

Hence,Option(B) $-\frac{1}{2}$ is the correct choice.

Comments

I think ans should be -1/2

check the step after taking x^2 comm.

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yes,Right.Thank You.