मराठी

Evaluate the following limits: limx→0[1+x3-1+xx] - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Evaluate the following limits: `lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`

बेरीज

उत्तर

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

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

Put 1 + x = y
As x → 0, y → 1

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

= `lim_(y -> 1)(y^(1/3) - y^(1/2))/(y - 1)`

= `lim_(y -> 1)((y^(1/3) - 1) - (y^(1/2) - 1))/(y - 1)`

= `lim_(y -> 1)((y^(1/3) - 1)/(y - 1) - (y^(1/2) - 1)/(y - 1))`

= `lim_(y -> 1) (y^(1/3) - 1^(1/3))/(y - 1) - lim_(y -> 1)(y^(1/2) - 1^(1/2))/(y - 1)`

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

= `1/3 - 1/2`

= `(2 - 3)/6`

= `-1/6`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Limits - EXERCISE 7.1 [पृष्ठ १००]

APPEARS IN

संबंधित प्रश्‍न

\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\] 


\[\lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x^3 - 3 x^2 + 2x} \right]\] 


\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]


\[\lim_{x \to \infty} \frac{\left( 3x - 1 \right) \left( 4x - 2 \right)}{\left( x + 8 \right) \left( x - 1 \right)}\] 


\[\lim_{n \to \infty} \left[ \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!} \right]\] 


\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . n^3}{\left( n - 1 \right)^4} \right]\] 


\[\lim_{x \to 0} \frac{\sin \left( 2 + x \right) - \sin \left( 2 - x \right)}{x}\]


\[\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^2}\] 


\[\lim_{x \to 0} \frac{\sin 2x \left( \cos 3x - \cos x \right)}{x^3}\] 


\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos 2x}\] 


\[\lim_{x \to 0} \frac{cosec x - \cot x}{x}\]


\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\] 


\[\lim_{x \to \frac{\pi}{8}} \frac{\cot 4x - \cos 4x}{\left( \pi - 8x \right)^3}\] 


\[\lim_{x \to 0^-} \frac{\sin \left[ x \right]}{\left[ x \right]} .\] 


\[\lim_{x \to 0}  \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]


Evaluate `lim_(h -> 0) ((a + h)^2 sin (a + h) - a^2 sina)/h`


Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`


Evaluate the following limit.

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


Evaluate the following limit:

`\underset{x->3}{lim}[sqrt(x +6)/(x)]`


Evaluate the Following limit:

`lim_(x->7)[[(root[3][x] - root[3][7])(root[3][x] + root[3][7])] / (x - 7)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×