Advertisements
Advertisements
Question
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . n^3}{\left( n - 1 \right)^4} \right]\]
Solution
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . + n^3}{\left( n - 1 \right)^4} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{\left[ \frac{n\left( n + 1 \right)}{2} \right]^2}{\left( n - 1 \right)^4} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{n^2 \left( n + 1 \right)^2}{4 \left( n - 1 \right)^4} \right]\]
Dividing the numerator and the denominator by n4:
\[\lim_{n \to \infty} \left[ \frac{\frac{n^2 \left( n + 1 \right)^2}{n^4}}{4\frac{\left( n - 1 \right)^4}{n^4}} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{\frac{\left( n + 1 \right)^2}{n^2}}{4 \left( \frac{n - 1}{n} \right)^4} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{\left( 1 + \frac{1}{n} \right)^2}{4 \left( 1 - \frac{1}{n} \right)^4} \right]\]
\[\text{ When } n \to \infty , \text{ then } \frac{1}{n} \to 0 . \]
\[\frac{\left( 1 + 0 \right)^2}{4 \left( 1 - 0 \right)^4}\]
\[ = \frac{1}{4}\]
APPEARS IN
RELATED QUESTIONS
\[\lim_{x \to - 5} \frac{2 x^2 + 9x - 5}{x + 5}\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
\[\lim_{x \to 5} \frac{x^3 - 125}{x^2 - 7x + 10}\]
\[\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\]
\[\lim_{x \to 1} \frac{x^4 - 3 x^3 + 2}{x^3 - 5 x^2 + 3x + 1}\]
\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\]
\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\]
\[\lim_{n \to \infty} \frac{n^2}{1 + 2 + 3 + . . . + n}\]
\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]
\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\]
\[\lim_{x \to 0} \frac{7x \cos x - 3 \sin x}{4x + \tan x}\]
\[\lim_{x \to 0} \frac{\sin \left( 2 + x \right) - \sin \left( 2 - x \right)}{x}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x}{\cos 2x - \cos 8x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}\]
\[\lim_{x \to 0} \frac{cosec x - \cot x}{x}\]
\[\lim_{x \to \pi} \frac{\sin x}{\pi - x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
\[\lim_{x \to \frac{\pi}{8}} \frac{\cot 4x - \cos 4x}{\left( \pi - 8x \right)^3}\]
\[\lim_{n \to \infty} \frac{\sin \left( \frac{a}{2^n} \right)}{\sin \left( \frac{b}{2^n} \right)}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to \frac{\pi}{6}} \frac{\cot^2 x - 3}{cosec x - 2}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\left( \frac{\pi}{2} - x \right) \sin x - 2 \cos x}{\left( \frac{\pi}{2} - x \right) + \cot x}\]
\[\lim_{x \to 0} \left( \cos x \right)^{1/\sin x}\]
\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
\[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]
\[\lim_{x \to 0^-} \frac{\sin x}{\sqrt{x}} .\]
\[\lim 2_{h \to 0} \left\{ \frac{\sqrt{3} \sin \left( \pi/6 + h \right) - \cos \left( \pi/6 + h \right)}{\sqrt{3} h \left( \sqrt{3} \cos h - \sin h \right)} \right\}\]
\[\lim_{n \to \infty} \frac{n!}{\left( n + 1 \right)! + n!}\] is equal to
\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] is equal to
\[\lim_{x \to 1} \left[ x - 1 \right]\] where [.] is the greatest integer function, is equal to
If \[f\left( x \right) = \begin{cases}\frac{\sin\left[ x \right]}{\left[ x \right]}, & \left[ x \right] \neq 0 \\ 0, & \left[ x \right] = 0\end{cases}\] where denotes the greatest integer function, then \[\lim_{x \to 0} f\left( x \right)\]
Evaluate the following limit:
`lim_(x -> 3) [sqrt(x + 6)/x]`
Evaluate the following limits: `lim_(y -> 1) [(2y - 2)/(root(3)(7 + y) - 2)]`
`lim_(x->3) (x^5 - 243)/(x^3 - 27)` = ?
`1/(ax^2 + bx + c)`
Evaluate the following limit :
`lim_(x->5)[(x^3-125)/(x^5-3125)]`
Evaluate the following limit:
`lim_(x->3)[(sqrt(x+6))/x]`
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]`
