Advertisements
Advertisements
Question
\[\lim_{n \to \infty} \left\{ \frac{1}{1 - n^2} + \frac{2}{1 - n^2} + . . . + \frac{n}{1 - n^2} \right\}\]
Options
(a) 0
(b) −1/2
(c) 1/2
(d) none of these
Advertisements
Solution
(b) \[\frac{- 1}{2}\]
\[\lim_{n \to \infty} \left[ \frac{1}{1 - n^2} + \frac{2}{1 - n^2} + . . . . . . + \frac{n}{1 - n^2} \right]\]
\[ = \lim_{n \to \infty} \left( \frac{1}{1 - n^2} \right) \left[ 1 + 2 + 3 . . . . . n \right]\]
\[ = \lim_{n \to \infty} \left( \frac{1}{1 - n^2} \right) \left[ \frac{n\left( n + 1 \right)}{2} \right]\]
\[ = \lim_{n \to \infty} \frac{1}{2}\left[ \frac{n\left( n + 1 \right)}{\left( 1 - n \right) \left( 1 + n \right)} \right]\]
\[ = \lim_{n \to \infty} \frac{1}{2} \frac{n}{1 - n}\]
\[ = \lim_{n \to \infty} \frac{1}{2} \frac{1}{\frac{1}{n} - 1}\]
\[ = \frac{1}{2}\left( - 1 \right)\]
\[ = \frac{- 1}{2}\]
APPEARS IN
RELATED QUESTIONS
\[\lim_{x \to 0} 9\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
\[\lim_{x \to 2} \left( \frac{x}{x - 2} - \frac{4}{x^2 - 2x} \right)\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{2}{x^2 - 2x} \right)\]
\[\lim_{x \to 27} \frac{\left( x^{1/3} + 3 \right) \left( x^{1/3} - 3 \right)}{x - 27}\]
\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\]
\[\lim_{x \to \infty} \sqrt{x^2 + cx - x}\]
\[\lim_{x \to \infty} \sqrt{x^2 + 7x - x}\]
\[\lim_{x \to \infty} \frac{x}{\sqrt{4 x^2 + 1} - 1}\]
\[\lim_{x \to \infty} \frac{\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2}}{\sqrt{x^2 + c^2} - \sqrt{x^2 + d^2}}\]
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . . n^3}{n^4} \right]\]
\[f\left( x \right) = \frac{a x^2 + b}{x^2 + 1}, \lim_{x \to 0} f\left( x \right) = 1\] and \[\lim_{x \to \infty} f\left( x \right) = 1,\]then prove that f(−2) = f(2) = 1
\[\lim_{x \to 0} \frac{\sin x \cos x}{3x}\]
\[\lim_{x \to 0} \frac{x^2 - \tan 2x}{\tan x}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\]
\[\lim_\theta \to 0 \frac{1 - \cos 4\theta}{1 - \cos 6\theta}\]
\[\lim_\theta \to 0 \frac{\sin 4\theta}{\tan 3\theta}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{1 - \sin x}\]
Evaluate the following limit:
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{1 - \cos6x}}{\sqrt{2}\left( \frac{\pi}{3} - x \right)}\]
\[\lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin 2\pi x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \sin 2x}{1 + \cos 4x}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{{cosec}^2 x - 2}{\cot x - 1}\]
\[\lim_{x \to 0^+} \left\{ 1 + \tan^2 \sqrt{x} \right\}^{1/2x}\]
\[\lim_{x \to 0} \left( \cos x \right)^{1/\sin x}\]
\[\lim_{x \to \pi/4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}\] is equal to
\[\lim_{x \to 1} \frac{\sin \pi x}{x - 1}\]
\[\lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x} - \sqrt{3}}}{x - 2}\] is equal to
Evaluate the following limit:
`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
Evaluate the following limits: `lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`
Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^2 - 25)]`
Evaluate the following Limit:
`lim_(x -> 0) ((1 + x)^"n" - 1)/x`
If f(x) = `{{:(1 if x "is rational"),(-1 if x "is rational"):}` is continuous on ______.
Evaluate the following limit:
`lim_(x->7)[((root(3)(x)-root(3)(7))(root(3)(x)+root(3)(7)))/(x-7)]`
Evaluate the following limit:
`\underset{x->3}{lim}[sqrt(x +6)/(x)]`
