मराठी

Evaluate: limx→1x4-xx-1

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

Given that `lim_(x -> 1) (x^4 - sqrt(x))/(sqrt(x) - 1)`

= `lim_(x -> 1) (sqrt(x)[(x)^(7/2) - 1])/(sqrt(x) - 1)`

= `lim_(x -> 1) (sqrt(x) ([x^(7/2) - (1)^(7/2)])/(x - 1))/(((x)^(1/2) - (1)^(1/2))/(x - 1))`  .....[Dividing the numerator and denominator of x – 1]

= `lim_(x -> 1) (((x)^(7/2) - (1)^(7/2))/(x - 1))/(((x)^(1/2) - (1)^(1/2))/(x - 1)) xx lim_(x -> 1) sqrt(x)`  .....`[because  lim_(x -> a) f(x) g(x) - lim_(x -> a) f(x) * lim_(x -> a) g(x)]`

= `(7/2 (1)^*7/2 - 1)/(1/2(1)^(1/2 - 1)) xx sqrt(1)`

= `(7/2)/(1/2)`

= 7

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Limits and Derivatives - Exercise [पृष्ठ २४०]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
पाठ 13 Limits and Derivatives
Exercise | Q 7 | पृष्ठ २४०

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Evaluate the following limit.

`lim_(x -> pi) (sin(pi - x))/(pi (pi - x))`


Evaluate the following limit.

`lim_(x → 0) x sec x`


Evaluate the following limit.

`lim_(x -> 0) (cosec x -  cot x)`


Evaluate the following limit.

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


Evaluate the following limit :

`lim_(x -> 0) [(x*tanx)/(1 - cosx)]`


Evaluate the following limit :

`lim_(x -> pi/6) [(2 - "cosec"x)/(cot^2x - 3)]`


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following limit :

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


Select the correct answer from the given alternatives.

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


Evaluate the following :

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


Evaluate the following :

`lim_(x -> pi/4) [(sinx - cosx)^2/(sqrt(2) - sinx - cosx)]`


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


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


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


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


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


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


Evaluate: `lim_(x -> 3) (x^2 - 9)/(x - 3)`


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


Evaluate: `lim_(x -> 1) (x^7 - 2x^5 + 1)/(x^3 - 3x^2 + 2)`


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


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


Evaluate: `lim_(x -> 0) (1 - cos mx)/(1 - cos nx)`


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


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


Evaluate: `lim_(x -> pi/6) (cot^2 x - 3)/("cosec"  x - 2)`


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


`lim_(x -> pi) (1 - sin  x/2)/(cos  x/2 (cos  x/4 - sin  x/4))`


`lim_(x -> pi) sinx/(x - pi)` is equal to ______.


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


`lim_(x -> 0) (1 - cos 4theta)/(1 - cos 6theta)` is ______.


`lim_(x -> pi/4) (sec^2x - 2)/(tan x - 1)` is equal to ______.


`lim_(x -> 1) ((sqrt(x) - 1)(2x - 3))/(2x^2 + x - 3)` is ______.


If `f(x) = {{:(x^2 - 1",", 0 < x < 2),(2x + 3",", 2 ≤ x < 3):}`, the quadratic equation whose roots are `lim_(x -> 2^-) f(x)` and `lim_(x -> 2^+) f(x)` is ______. 


The value of `lim_(x → ∞) ((x^2 - 1)sin^2(πx))/(x^4 - 2x^3 + 2x - 1)` 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_(x rightarrow π/2) ([1 - tan (x/2)] (1 - sin x))/([1 + tan (x/2)] (π - 2x)^3` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×