मराठी

Lim X → 0 Log ( 3 + X ) − Log ( 3 − X ) X - Mathematics

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

\[\lim_{x \to 0} \left[ \frac{\log \left( 3 + x \right) - \log \left( 3 - x \right)}{x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\log \left( \frac{3 + x}{3 - x} \right)}{x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\log \left( 1 + \frac{3 + x}{3 - x} - 1 \right)}{x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\log \left( 1 + \frac{3 + x - 3 + x}{3 - x} \right)}{x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\log \left( 1 + \frac{2x}{3 - x} \right)}{x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\log \left( 1 + \frac{2x}{3 - x} \right)}{\left( \frac{2x}{3 - x} \right) \times \frac{3 - x}{2}} \right]\]
\[ = \frac{1 \times 2}{3 - 0}\]
\[ = \frac{2}{3}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.1 [पृष्ठ ७१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.1 | Q 23 | पृष्ठ ७१

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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


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


\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]


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


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


If \[\lim_{x \to 3} \frac{x^n - 3^n}{x - 3} = 108,\]  find the value of n


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


\[\lim_{x \to \infty} \sqrt{x^2 + 7x - x}\] 


\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\] 


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


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


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


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


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


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


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


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


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


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


If  \[\lim_{x \to 0} kx  cosec x = \lim_{x \to 0} x  cosec kx,\] 


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


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


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


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


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


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


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


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


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


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


The value of \[\lim_{x \to 0} \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} - \sqrt{a - x}}\] 


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


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


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


Evaluate the following limits: if `lim_(x -> 1)[(x^4 - 1)/(x - 1)] = lim_(x -> "a") [(x^3 - "a"^3)/(x - "a")]`, find all the value of a.


Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`


Evaluate the following limits: `lim_(x -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`


`lim_(x->3) (x^5 - 243)/(x^3 - 27)` = ?


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×