हिंदी

Find the positive integer n so that limx→3xn-3nx-3 = 108.

Advertisements
Advertisements

प्रश्न

Find the positive integer n so that `lim_(x -> 3) (x^n - 3^n)/(x - 3)` = 108.

योग
Advertisements

उत्तर

We have `lim_(x -> 3) (x^n - 3^n)/(x - 3) = n(3)^(n - 1)`

Therefore, `n(3)^(n - 1)` = 108

= 4(27)

= `4(3)^(4 - 1)`

Comparing, we get

n = 4 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 13: Limits and Derivatives - Solved Examples [पृष्ठ २२८]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
अध्याय 13 Limits and Derivatives
Solved Examples | Q 3 | पृष्ठ २२८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Evaluate the following limit.

`lim_(x → 0) x sec x`


Evaluate the following limit.

`lim_(x -> 0) (sin ax + bx)/(ax + sin bx) a, b, a+ b != 0`


Evaluate the following limit :

`lim_(theta -> 0) [(sin("m"theta))/(tan("n"theta))]`


Evaluate the following limit :

`lim_(theta -> 0) [(1 - cos2theta)/theta^2]`


Evaluate the following limit :

`lim_(x ->0)((secx - 1)/x^2)`


Evaluate the following limit :

`lim_(x -> 0) [(cos("a"x) - cos("b"x))/(cos("c"x) - 1)]`


Evaluate the following limit :

`lim_(x -> pi/4) [(tan^2x - cot^2x)/(secx - "cosec"x)]`


Select the correct answer from the given alternatives.

`lim_(x → π/3) ((tan^2x - 3)/(sec^3x - 8))` =


Select the correct answer from the given alternatives.

`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]` =


`lim_{x→0}((3^x - 3^xcosx + cosx - 1)/(x^3))` is equal to ______ 


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


Evaluate `lim_(x -> pi/2) (secx - tanx)`


Evaluate `lim_(x -> 0) (tanx - sinx)/(sin^3x)`


Evaluate `lim_(x -> 0) (cos ax - cos bx)/(cos cx - 1)`


`lim_(x -> 0) |x|/x` is equal to ______.


Evaluate: `lim_(x -> 1/2) (4x^2 - 1)/(2x  - 1)`


Evaluate: `lim_(x -> 0) (sqrt(1 + x^3) - sqrt(1 - x^3))/x^2`


Evaluate: `lim_(x -> 1/2) (8x - 3)/(2x - 1) - (4x^2 + 1)/(4x^2 - 1)`


Evaluate: `lim_(x -> 0) (1 - cos 2x)/x^2`


Evaluate: `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`


Evaluate: `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`


Evaluate: `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`


cos (x2 + 1)


`(ax + b)/(cx + d)`


`x^(2/3)`


x cos x


`lim_(y -> 0) ((x + y) sec(x + y) - x sec x)/y`


`lim_(x -> 0) ((1 + x)^n - 1)/x` is equal to ______.


`lim_(x -> 0) ("cosec" x - cot x)/x` is equal to ______.


If `f(x) = {{:(sin[x]/[x]",", [x] ≠ 0),(0",", [x] = 0):}`, where [.] denotes the greatest integer function, then `lim_(x -> 0) f(x)` is equal to ______.


`lim_(x -> 0) |sinx|/x` is ______.


If `f(x) = tanx/(x - pi)`, then `lim_(x -> pi) f(x)` = ______.


`lim_(x -> 0) (sin mx cot  x/sqrt(3))` = 2, then m = ______. 


`lim_(x -> 3^+) x/([x])` = ______.


Let Sk = `sum_(r = 1)^k tan^-1(6^r/(2^(2r + 1) + 3^(2r + 1)))`. Then `lim_(k→∞)` Sk = is equal to ______.


If L = `lim_(x→∞)(x^2sin  1/x - x)/(1 - |x|)`, then value of L is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×