हिंदी

Evaluate the following limit : limx→1[x+x2+x3+.........+xn-nx-1] - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Evaluate the following limit :

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

योग
Advertisements

उत्तर

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

= `lim_(x -> 1)[(x + x^2 + x^3 + .... +  x^"n" - (1 + 1 + 1 + ...  "n times"))/(x - 1)]`

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

= `lim_(x -> 1)[(x^1 - 1^1)/(x - 1) + (x^2 - 1^2)/(x - 1) + (x^3 - 1^3)/(x - 1) + ... +  (x^"n" - 1^"n")/(x - 1)]`

= 1 (1)0 + 2 (1)1 + 3 (1)2 + 4 (1)3 + ... + n (1)n–1  ...`[lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`

= 1 + 2 + 3 + 4 + ... + n

= `("n"("n" + 1))/2`

shaalaa.com
Concept of Limits
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Limits - Exercise 7.1 [पृष्ठ १३९]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 7 Limits
Exercise 7.1 | Q III. (1) | पृष्ठ १३९

संबंधित प्रश्न

Evaluate the following limit:

`lim_(z -> -3) [sqrt("z" + 6)/"z"]`


Evaluate the following limit:

`lim_(z -> -5)[((1/z + 1/5))/(z + 5)]`


Evaluate the following limit:

`lim_(x -> 3)[sqrt(2x + 6)/x]`


Evaluate the following limit:

If `lim_(x -> 1)[(x^4 - 1)/(x - 1)]` = `lim_(x -> "a")[(x^3 - "a"^3)/(x - "a")]`, find all possible values of a


Evaluate the following limit :

`lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`


In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.

`lim_(x -> -3) (3x + 2)` = – 7


Evaluate the following :

Given that 7x ≤ f(x) ≤ 3x2 – 6 for all x. Determine the value of `lim_(x -> 3) "f"(x)`


In problems 1 – 6, using the table estimate the value of the limit.

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

x 1.9 1.99 1.999 2.001 2.01 2.1
f(x) 0.344820 0.33444 0.33344 0.333222 0.33222 0.332258

In problems 1 – 6, using the table estimate the value of the limit
`lim_(x -> 0) (sqrt(x + 3) - sqrt(3))/x`

x – 0.1  – 0.01 – 0.001 0.001 0.01 0.1
f(x) 0.2911 0.2891 0.2886 0.2886 0.2885 0.28631

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

x – 0.1  – 0.01 – 0.001 0.0001 0.01 0.1
f(x) 0.04995 0.0049999 0.0004999 – 0.0004999 – 0.004999 – 0.04995

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 -> 0) sec x`


Sketch the graph of f, then identify the values of x0 for which `lim_(x -> x_0)` f(x) exists.

f(x) = `{{:(x^2",", x ≤ 2),(8 - 2x",", 2 < x < 4),(4",", x ≥ 4):}`


Evaluate the following limits:

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


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> 1) (root(3)(7 + x^3) - sqrt(3 + x^2))/(x - 1)`


Evaluate the following limits:

`lim_(x -> 5) (sqrt(x - 1) - 2)/(x - 5)`


Find the left and right limits of f(x) = `(x^2 - 4)/((x^2 + 4x+ 4)(x + 3))` at x = – 2


Evaluate the following limits:

`lim_(x -> oo) (x^4 - 5x)/(x^2 - 3x + 1)`


Evaluate the following limits:

`lim_(x -> oo) (1 + x - 3x^3)/(1 + x^2 +3x^3)`


Evaluate the following limits:

`lim_(x -> oo)(1 + 1/x)^(7x)`


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 -> oo) ((2x^2 + 3)/(2x^2 + 5))^(8x^2 + 3)`


Evaluate the following limits:

`lim_(x -> 0) (sin^3(x/2))/x^2`


Evaluate the following limits:

`lim_(x -> 0) (2 "arc"sinx)/(3x)`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> oo) x [3^(1/x) + 1 - cos(1/x) - "e"^(1/x)]`


Evaluate the following limits:

`lim_(x - oo){x[log(x + "a") - log(x)]}`


Evaluate the following limits:

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


Evaluate the following limits:

`lim_(x -> oo) ((x^2 - 2x + 1)/(x^2 -4x + 2))^x`


Choose the correct alternative:

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


Choose the correct alternative:

`lim_(x -> 0) ("a"^x - "b"^x)/x` =


Choose the correct alternative:

If `lim_(x -> 0) (sin "p"x)/(tan 3x)` = 4, then the value of p is


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


If `lim_(x -> 1) (x + x^2 + x^3|+ .... + x^n - n)/(x - 1)` = 820, (n ∈ N) then the value of n is equal to ______.


`lim_(x→-1) (x^3 - 2x - 1)/(x^5 - 2x - 1)` = ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×