मराठी

Lim N → ∞ [ 1 3 + 2 3 + . . . . N 3 N 4 ] - Mathematics

Advertisements
Advertisements

प्रश्न

\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . . n^3}{n^4} \right]\]

Advertisements

उत्तर

\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . + n^3}{n^4} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{\left[ \frac{n\left( n + 1 \right)}{2} \right]^2}{n^4} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{n^2 \left( n + 1 \right)^2}{4 n^4} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{n^2 \left( n + 1 \right)^2}{4 n^4} \right]\]
\[ = \lim_{n \to \infty} \left[ \frac{\left( n + 1 \right)^2}{4 n^2} \right]\]
\[ = \lim_{n \to \infty} \left[ \left( \frac{n + 1}{n} \right)^2 \times \frac{1}{4} \right]\]
\[ = \lim_{n \to \infty} \left[ \left( 1 + \frac{1}{n} \right)^2 \times \frac{1}{4} \right]\]
\[\text{ When } n \to \infty , \text{ then } \frac{1}{n} \to 0 . \]
\[ \Rightarrow \frac{1}{4}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.6 [पृष्ठ ३९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.6 | Q 16 | पृष्ठ ३९

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\] 


\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\] 


\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{2}{x^2 - 2x} \right)\] 


\[\lim_{x \to 0} \frac{\left( a + x \right)^2 - a^2}{x}\] 


\[\lim_{x \to 3} \left( x^2 - 9 \right) \left[ \frac{1}{x + 3} + \frac{1}{x - 3} \right]\] 


\[\lim_{x \to 1} \left\{ \frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3 x^2 + 2x} \right\}\] 


\[\lim_{x \to a} \frac{\left( x + 2 \right)^{3/2} - \left( a + 2 \right)^{3/2}}{x -  a}\]


\[\lim_{x \to a} \frac{x^{2/3} - a^{2/3}}{x^{3/4} - a^{3/4}}\] 


\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . n^3}{\left( n - 1 \right)^4} \right]\] 


\[f\left( x \right) = \frac{a x^2 + b}{x^2 + 1}, \lim_{x \to 0} f\left( x \right) = 1\] and \[\lim_{x \to \infty} f\left( x \right) = 1,\]then prove that f(−2) = f(2) = 1


\[\lim_{x \to 0} \frac{2x - \sin x}{\tan x + x}\] 


\[\lim_{x \to 0} \frac{\sin 3x - \sin x}{\sin x}\] 


\[\lim_{x \to 0} \frac{\tan 3x - 2x}{3x - \sin^2 x}\] 


\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin \left( a + h \right) - a^2 \sin a}{h}\] 


\[\lim_{x \to 0} \frac{1 - \cos 2x + \tan^2 x}{x \sin x}\] 


\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\] 


\[\lim_{x \to 0} \frac{x \cos x + \sin x}{x^2 + \tan x}\] 


\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]


\[\lim_{x \to \frac{\pi}{2}} \frac{\sin 2x}{\cos x}\] 


\[\lim_{x \to \frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \tan x\]


\[\lim_{x \to 0} \left( \cos x \right)^{1/\sin x}\] 


Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]


Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]


Write the value of \[\lim_{x \to 0^-} \frac{\sin \left[ x \right]}{\left[ x \right]} .\]


\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to 


\[\lim_{n \to \infty} \frac{n!}{\left( n + 1 \right)! + n!}\]  is equal to


\[\lim_\theta \to \pi/2 \frac{1 - \sin \theta}{\left( \pi/2 - \theta \right) \cos \theta}\] is equal to 


The value of \[\lim_{x \to \infty} \frac{\left( x + 1 \right)^{10} + \left( x + 2 \right)^{10} + . . . + \left( x + 100 \right)^{10}}{x^{10} + {10}^{10}}\] is 


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 -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`


Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`


Number of values of x where the function

f(x) = `{{:((tanxlog(x - 2))/(x^2 - 4x + 3); x∈(2, 4) - {3, π}),(1/6tanx; x = 3","  π):}`

is discontinuous, is ______.


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->5)[(x^3-125)/(x^5-3125)]`


Evaluate the following limits: `lim_(x -> 3) [sqrt(x + 6)/x]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×