मराठी

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.

बेरीज
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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Limits and Derivatives - Solved Examples [पृष्ठ २२८]

APPEARS IN

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

व्हिडिओ 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_(x ->0)((secx - 1)/x^2)`


Evaluate the following limit :

`lim_(x -> pi/4) [(cosx - sinx)/(cos2x)]`


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]` =


Evaluate the following :

`lim_(x -> 0)[(secx^2 - 1)/x^4]`


Evaluate the following :

`lim_(x -> "a") [(sinx - sin"a")/(x - "a")]`


`lim_{x→-5} (sin^-1(x + 5))/(x^2 + 5x)` is equal to ______ 


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


Evaluate `lim_(x -> 0) (sqrt(2 + x) - sqrt(2))/x`


Evaluate `lim_(x -> 0)  (sin(2 + x) - sin(2 - x))/x`


Evaluate `lim_(x -> pi/6) (2sin^2x + sin x - 1)/(2sin^2 x - 3sin x + 1)`


Evaluate `lim_(x -> a) (sqrt(a + 2x) - sqrt(3x))/(sqrt(3a + x) - 2sqrt(x))`


Find the derivative of f(x) = `sqrt(sinx)`, by first principle.


`lim_(x -> pi/2) (1 - sin x)/cosx` is equal to ______.


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


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


Evaluate: `lim_(x -> a) ((2 + x)^(5/2) - (a + 2)^(5/2))/(x - a)`


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


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


Evaluate: `lim_(x -> 3) (x^3 + 27)/(x^5 + 243)`


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


Evaluate: `lim_(x -> a) (sin x - sin a)/(sqrt(x) - sqrt(a))`


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


`lim_(x -> 0) ((sin(alpha + beta) x + sin(alpha - beta)x + sin 2alpha x))/(cos 2betax - cos 2alphax) * x`


`lim_(x -> 0) (x^2 cosx)/(1 - cosx)` is ______.


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


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


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


If `lim_(x→∞) 1/(x + 1) tan((πx + 1)/(2x + 2)) = a/(π - b)(a, b ∈ N)`; then the value of a + b is ______.


`lim_(x rightarrow ∞) sum_(x = 1)^20 cos^(2n) (x - 10)` is equal to ______.


`lim_(theta → -pi/4) (cos theta + sin theta)/(theta + pi/4)` =


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×