Advertisements
Advertisements
प्रश्न
Evaluate: \[\lim_{n \to \infty} \frac{1 . 2 + 2 . 3 + 3 . 4 + . . . + n\left( n + 1 \right)}{n^3}\]
Advertisements
उत्तर
\[\lim_{n \to \infty} \frac{1 . 2 + 2 . 3 + 3 . 4 + . . . + n\left( n + 1 \right)}{n^3}\]
\[ = \lim_{n \to \infty} \frac{\sum^n_{k = 1} k\left( k + 1 \right)}{n^3}\]
\[ = \lim_{n \to \infty} \frac{\sum^n_{k = 1} k^2 + \sum^n_{k = 1} k}{n^3}\]
\[ = \lim_{n \to \infty} \frac{\frac{n\left( n + 1 \right)\left( 2n + 1 \right)}{6} + \frac{n\left( n + 1 \right)}{2}}{n^3}\]
\[ = \lim_{n \to \infty} \frac{2n\left( n + 1 \right)\left( n + 2 \right)}{6 n^3}\]
\[ = \frac{1}{3} \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)\left( 1 + \frac{2}{n} \right)\]
\[ = \frac{1}{3} \times \left( 1 + 0 \right) \times \left( 1 + 0 \right) \left( \lim_{n \to \infty} \frac{1}{n} = 0 \right)\]
\[ = \frac{1}{3}\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\]
\[\lim_{x \to 0} \frac{x^{2/3} - 9}{x - 27}\]
\[\lim_{x \to - 1}{\left( 4 x^2 + 2 \right)}\]
\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]
\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\]
\[\lim_{x \to \sqrt{3}} \frac{x^2 - 3}{x^2 + 3 \sqrt{3}x - 12}\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to 0} \frac{\left( a + x \right)^2 - a^2}{x}\]
\[\lim_{x \to 1} \left\{ \frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3 x^2 + 2x} \right\}\]
\[\lim_{x \to a} \frac{\left( x + 2 \right)^{5/2} - \left( a + 2 \right)^{5/2}}{x - a}\]
\[\lim_{x \to a} \frac{x^{5/7} - a^{5/7}}{x^{2/7} - a^{2/7}}\]
\[\lim_{x \to a} \frac{x^{2/3} - a^{2/3}}{x^{3/4} - a^{3/4}}\]
\[\lim_{x \to \infty} \frac{\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2}}{\sqrt{x^2 + c^2} - \sqrt{x^2 + d^2}}\]
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
\[\lim_{x \to 0} \frac{\sin 3x}{5x}\]
\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\]
\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\]
\[\lim_{x \to 0} \frac{\sin^2 4 x^2}{x^4}\]
\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\]
\[\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}\]
\[\lim_{n \to \infty} \frac{\sin \left( \frac{a}{2^n} \right)}{\sin \left( \frac{b}{2^n} \right)}\]
\[\lim_{x \to - 1} \frac{x^2 - x - 2}{\left( x^2 + x \right) + \sin \left( x + 1 \right)}\]
\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( 4x - \pi \right)^2}\]
\[\lim_{x \to \frac{\pi}{6}} \frac{\cot^2 x - 3}{cosec x - 2}\]
\[\lim_{x \to \frac{3\pi}{2}} \frac{1 + {cosec}^3 x}{\cot^2 x}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log a}{x}\]
\[\lim_{x \to 0} \frac{8^x - 2^x}{x}\]
\[\lim_{x \to 0} \frac{\sin x^0}{x}\]
\[\lim_{x \to 1} \frac{\sin \pi x}{x - 1}\]
The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\] is
\[\lim_{x \to 3} \frac{\sum^n_{r = 1} x^r - \sum^n_{r = 1} 3^r}{x - 3}\]is real to
\[\lim_{x \to 1} \left[ x - 1 \right]\] where [.] is the greatest integer function, 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 limit:
`lim_(x->3)[sqrt(x+6)/x]`
Evaluate the following limit :
`lim_(x->5)[(x^3-125)/(x^5-3125)]`
Evaluate the Following limit:
`lim_(x->7)[[(root[3][x] - root[3][7])(root[3][x] + root[3][7])] / (x - 7)]`
