हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान कक्षा ११

Choose the correct alternative: nnnnnlimx→∞(1n2+2n2+3n2+...+nn2) is

Advertisements
Advertisements

प्रश्न

Choose the correct alternative:

`lim_(x -> oo) (1/"n"^2 + 2/"n"^2 + 3/"n"^2 + ... + "n"/"n"^2)` is

विकल्प

  • `1/2`

  • 0

  • 1

  • `oo`

MCQ
Advertisements

उत्तर

`1/2`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.6 [पृष्ठ १३०]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.6 | Q 16 | पृष्ठ १३०

संबंधित प्रश्न

Evaluate the following limit:

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


Evaluate the following :

`lim_(x -> 0) {1/x^12 [1 - cos(x^2/2) - cos(x^4/4) + cos(x^2/2) cos(x^4/4)]}`


In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) sin x/x`

x – 0.1  – 0.01 – 0.001 0.001 0.01 0.1
f(x) 0.99833 0.99998 0.99999 0.99999 0.99998 0.99833

In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?

`lim_(x -> x/2) tan x`


Write a brief description of the meaning of the notation `lim_(x -> 8) f(x)` = 25


Evaluate the following limits:

`lim_(x ->) (x^"m" - 1)/(x^"n" - 1)`, m and n are integers


Evaluate the following limits:

`lim_(x -> 1) (sqrt(x) - x^2)/(1 - sqrt(x))`


Evaluate the following limits:

`lim_(x -> 2) (2 - sqrt(x + 2))/(root(3)(2) - root(3)(4 - x))`


Evaluate the following limits:

`lim_(x - 0) (sqrt(1 + x^2) - 1)/x`


Find the left and right limits of f(x) = tan x at x = `pi/2`


Evaluate the following limits:

`lim_(x -> 0)(1 + x)^(1/(3x))`


Evaluate the following limits:

`lim_(x -> oo)(1 + "k"/x)^("m"/x)`


Evaluate the following limits:

`lim_(x -> 0) (sqrt(1 + sinx) - sqrt(1 - sinx))/tanx`


Choose the correct alternative:

`lim_(x - oo) sqrt(x^2 - 1)/(2x + 1)` =


Choose the correct alternative:

`lim_(x -> 3) [x]` =


Choose the correct alternative:

`lim_(x -> 0) (x"e"^x - sin x)/x` is


Choose the correct alternative:

`lim_(alpha - pi/4) (sin alpha - cos alpha)/(alpha - pi/4)` is


Choose the correct alternative:

`lim_(x -> 0) ("e"^tanx - "e"^x)/(tan x - x)` =


`lim_(x -> 5) |x - 5|/(x - 5)` = ______.


`lim_(x→∞)((x + 7)/(x + 2))^(x + 4)` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×