English

Lim X → − 1 X 3 − 3 X + 1 X − 1

Advertisements
Advertisements

Question

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

Advertisements

Solution

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.2 [Page 18]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.2 | Q 11 | Page 18

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

\[\lim_{x \to 0} 9\] 


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


\[\lim_{x \to 0} \frac{3x + 1}{x + 3}\] 


\[\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\] 


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


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


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


\[\lim_{x \to 1} \frac{\sqrt{x^2 - 1} + \sqrt{x - 1}}{\sqrt{x^2 - 1}}, x > 1\] 


\[\lim_{x \to a} \frac{\left( x + 2 \right)^{5/2} - \left( a + 2 \right)^{5/2}}{x - a}\] 


\[\lim_{x \to \infty} \frac{x}{\sqrt{4 x^2 + 1} - 1}\] 


`lim_(x->∞) [x{sqrt(x^2+1) - sqrt(x^2-1)}]`


\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\] 


\[\lim_{x \to 0} \left[ \frac{x^2}{\sin x^2} \right]\] 


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


\[\lim_{x \to 0} \frac{3 \sin x - 4 \sin^3 x}{x}\] 


\[\lim_{x \to 0} \frac{\sin^2 4 x^2}{x^4}\] 


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


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


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


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


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


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


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


\[\lim_\theta \to 0 \frac{\sin 4\theta}{\tan 3\theta}\] 


\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]


Evaluate the following limit: 

\[\lim_{x \to 0} \frac{\sin\left( \alpha + \beta \right)x + \sin\left( \alpha - \beta \right)x + \sin2\alpha x}{\cos^2 \beta x - \cos^2 \alpha x}\]


\[\lim_{x \to \frac{\pi}{2}} \frac{\sin 2x}{\cos x}\] 


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


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


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


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


Evaluate the following limit:

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


Evaluate the following Limit:

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


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


If `f(x) = {{:(x + 2",",  x ≤ - 1),(cx^2",", x > -1):}`, find 'c' if `lim_(x -> -1) f(x)` exists


Evaluate the following limit :

`lim_(x->7)[[(root3(x)- root3(7))(root3(x) + root3(7)))/(x-7)]`


Evaluate the following limit:

`lim_(x->3)[(sqrt(x+6))/x]`


Evaluate the following limit:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×