मराठी

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

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

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

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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


\[\lim_{x \to 0} 9\] 


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


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


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


\[\lim_{x \to 1} \frac{1 - x^{- 1/3}}{1 - x^{- 2/3}}\] 


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


\[\lim_{x \to 27} \frac{\left( x^{1/3} + 3 \right) \left( x^{1/3} - 3 \right)}{x - 27}\] 


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


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


\[\lim_\theta \to 0 \frac{\sin 3\theta}{\tan 2\theta}\] 


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


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


\[\lim_\theta \to 0 \frac{1 - \cos 4\theta}{1 - \cos 6\theta}\] 


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


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


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


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


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


\[\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n\]


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


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

 

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


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


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


\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] 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 


\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] is equal to 


\[\lim_{x \to 0} \frac{8}{x^8}\left\{ 1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cos \frac{x^2}{4} \right\}\] is equal to 


\[\lim_{x \to \infty} \frac{\left| x \right|}{x}\]  is equal to 


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


Evaluate `lim_(h -> 0) ((a + h)^2 sin (a + h) - a^2 sina)/h`


`1/(ax^2 + bx + c)`


Number of values of x where the function

f(x) = `{{:((tanxlog(x - 2))/(x^2 - 4x + 3); x∈(2, 4) - {3, π}),(1/6tanx; x = 3","  π):}`

is discontinuous, is ______.


If f(x) = `{{:(1 if x  "is rational"),(-1 if x  "is rational"):}` is continuous on ______.


Evaluate the following limits: `lim_(x -> 3) [sqrt(x + 6)/x]`


Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`


Evaluate the following limit.

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×