English

Lim X → 3 ( X 2 − 9 ) [ 1 X + 3 + 1 X − 3 ]

Advertisements
Advertisements

Question

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

Advertisements

Solution

\[\lim_{x \to 3} \left[ \left( x^2 - 9 \right)\left\{ \frac{1}{x + 3} + \frac{1}{x - 3} \right\} \right]\]
\[ = \lim_{x \to 3} \left[ \left( x^2 - 9 \right)\left\{ \frac{x - 3 + x + 3}{\left( x + 3 \right)\left( x - 3 \right)} \right\} \right]\]
\[ = \lim_{x \to 3} \left[ \left( x^2 - 9 \right)\left( \frac{2x}{x^2 - 9} \right) \right]\]
\[ = \lim_{x \to 3} \left( 2x \right)\]
\[ = 2 \times 3\]
\[ = 6\]

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 24 | Page 23

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Show that \[\lim_{x \to 0} \frac{x}{\left| x \right|}\] does not exist.


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


\[\lim_{x \to 1} \frac{\sqrt{x + 8}}{\sqrt{x}}\] 


\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\] 


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


\[\lim_{x \to 1/4} \frac{4x - 1}{2\sqrt{x} - 1}\] 


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


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


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


If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = \lim_{x \to 5} \left( 4 + x \right),\] find all possible values of a


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


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


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


\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\] 


\[\lim_{x \to 0} \frac{7x \cos x - 3 \sin x}{4x + \tan x}\] 


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


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


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


\[\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}\] 


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


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


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


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


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


\[\lim_{n \to \infty} 2^{n - 1} \sin \left( \frac{a}{2^n} \right)\] 

 


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


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


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


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


\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\]  is equal at 


\[\lim_{n \to \infty} \left\{ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right\}\]is equal to


The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\]  is


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


\[\lim_{x \to \pi/3} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\] is equal to 


If \[f\left( x \right) = \left\{ \begin{array}{l}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0\end{array}, \right.\] then \[\lim_{x \to 0} f\left( x \right)\]  equals 


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


Evaluate the following limits: if `lim_(x -> 5)[(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.


Evaluate the following limits: `lim_(y -> 1) [(2y - 2)/(root(3)(7 + y) - 2)]`


if `lim_(x -> 2) (x^"n"- 2^"n")/(x - 2)` = 80 then find the value of n.


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×