English

Lim X → a X 2 / 7 − a 2 / 7 X − a

Advertisements
Advertisements

Question

\[\lim_{x \to a} \frac{x^{2/7} - a^{2/7}}{x - a}\] 

Advertisements

Solution

\[\lim_{x \to a} \left[ \frac{x^\frac{2}{7} - a^\frac{2}{7}}{x - a} \right]\]
\[ = \frac{2}{7} a^\frac{2}{7} - 1 \left[ \because \lim_{x \to a} \left[ \frac{x^n - a^n}{x - a} \right] = n a^{n - 1} \right]\]
\[ = \frac{2}{7} a^{- \frac{5}{7}}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.5 [Page 33]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.5 | Q 4 | Page 33

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

\[\lim_{x \to - 1}{\left( 4 x^2 + 2 \right)}\]


\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\] 


\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]


\[\lim_{x \to 3} \frac{x^4 - 81}{x^2 - 9}\] 


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


\[\lim_{x \to 2} \frac{x^4 - 16}{x - 2}\] 


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


\[\lim_{x \to 3} \left( x^2 - 9 \right) \left[ \frac{1}{x + 3} + \frac{1}{x - 3} \right]\] 


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


Evaluate the following limit:

\[\lim_{x \to 1} \frac{x^7 - 2 x^5 + 1}{x^3 - 3 x^2 + 2}\] 


\[\lim_{x \to \infty} \frac{3 x^3 - 4 x^2 + 6x - 1}{2 x^3 + x^2 - 5x + 7}\] 


\[\lim_{x \to \infty} \frac{\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2}}{\sqrt{x^2 + c^2} - \sqrt{x^2 + d^2}}\] 


\[\lim_{x \to - \infty} \left( \sqrt{x^2 - 8x} + x \right)\] 


\[\lim_{x \to 0} \frac{\sin 5x - \sin 3x}{\sin x}\] 


\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]


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


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


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


\[\lim_{x \to 0} \frac{3 \sin^2 x - 2 \sin x^2}{3 x^2}\] 


\[\lim_{x \to 0} \frac{1 - \cos 2x}{3 \tan^2 x}\] 


\[\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}\]


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


\[\lim_{x \to 1} \frac{1 - x^2}{\sin 2\pi x}\] 


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


\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{1 - \sqrt{2} \sin x}\] 


\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\] 


\[\lim_{x \to \frac{\pi}{4}} \frac{\cos x - \sin x}{\left( \frac{\pi}{4} - x \right) \left( \cos x + \sin x \right)}\]


\[\lim_{x \to \frac{\pi}{4}} \frac{2 - {cosec}^2 x}{1 - \cot x}\] 


\[\lim_{x \to 0} \left( \cos x \right)^{1/\sin x}\] 


\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\] is equal to 


\[\lim_{x \to 1} \left[ x - 1 \right]\] where [.] is the greatest integer function, is equal to 


Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^2 - 25)]`


Evaluate the following Limit:

`lim_(x -> 0) ((1 + x)^"n" - 1)/x`


If the value of `lim_(x -> 1) (1 - (1 - x))^"m"/x` is 99, then n = ______.


If `lim_(x -> 1) (x^4 - 1)/(x - 1) = lim_(x -> k) (x^3 - l^3)/(x^2 - k^2)`, then find the value of k.


Evaluate the following limit :

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


Evaluate the following limit.

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


Evaluate the Following limit: 

`lim_ (x -> 3) [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×