English

P Lim X → 2 X 3 − 8 X 2 − 4

Advertisements
Advertisements

Question

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

Advertisements

Solution

\[\lim_{x \to 2} \left[ \frac{x^3 - 8}{x^2 - 4} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 2} \left[ \frac{\left( x - 2 \right)\left( x^2 + 2x + 4 \right)}{\left( x - 2 \right)\left[ x + 2 \right]} \right] \binom{ \because A^3 - B^3 = \left( A - B \right)\left( A^2 + AB + B^2 \right)}{ \because A^2 - B^2 = \left( A - B \right)\left( A + B \right)}\]
\[ = \frac{2^2 + 2 \times 2 + 4}{2 + 2}\]
\[ = \frac{12}{4}\]
\[ = 3\]

 

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.3 [Page 23]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.3 | Q 4 | Page 23

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


\[\lim_{x \to 0} \frac{2 x^2 + 3x + 4}{x^2 + 3x + 2}\] 


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


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


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


\[\lim_{x \to 5} \frac{x^3 - 125}{x^2 - 7x + 10}\] 


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


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


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


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


\[\lim_{x \to \infty} \left[ \left\{ \sqrt{x + 1} - \sqrt{x} \right\} \sqrt{x + 2} \right]\] 


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


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


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


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


\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]


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


Evaluate the following limit: 

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


Evaluate the following limits: 

\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - 1}\] 


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


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


\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\] 


\[\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a}\] 


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


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


\[\lim_{x \to \frac{\pi}{6}} \frac{\cot^2 x - 3}{cosec x - 2}\]


Evaluate the following limit:

\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]

 


\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\] 


\[\lim_{x \to 0^+} \left\{ 1 + \tan^2 \sqrt{x} \right\}^{1/2x}\]


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


Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\] 


Write the value of \[\lim_{x \to \infty} \frac{\sin x}{x} .\] 


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


\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to


\[\lim_{x \to 3} \frac{\sum^n_{r = 1} x^r - \sum^n_{r = 1} 3^r}{x - 3}\]is real to


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


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


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×