मराठी

Lim X → 0 Sin a X + B X a X + Sin B X - Mathematics

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

\[\lim_{x \to 0} \left[ \frac{\sin \left( ax \right) + bx}{ax + \sin \left( bx \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\frac{\sin ax}{ax} \times ax + bx}{ax + \frac{\sin bx}{bx} \times bx} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( \frac{\sin ax}{ax} \times a + b \right)x}{\left( a + \frac{\sin bx}{bx} \times b \right)x} \right]\]
\[ = \frac{1 \times a + b}{a + b}\]
\[ = 1\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.7 [पृष्ठ ५१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.7 | Q 58 | पृष्ठ ५१

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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


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


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


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


Evaluate the following limit:

\[\lim_{x \to 1} \frac{x^7 - 2 x^5 + 1}{x^3 - 3 x^2 + 2}\] 


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


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


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


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


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


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


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


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


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


\[\lim_{x \to 0} \left( cosec x - \cot x \right)\]


Evaluate the following limit:

\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{1 - \cos6x}}{\sqrt{2}\left( \frac{\pi}{3} - x \right)}\]


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


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


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


\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{1 - \sqrt{2} \sin x}\] 


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


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


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


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


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


Write the value of \[\lim_{x \to 2} \frac{\left| x - 2 \right|}{x - 2} .\] 


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


\[\lim_{x \to 0}  \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]


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


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


\[\lim_{x \to \infty} a^x \sin \left( \frac{b}{a^x} \right), a, b > 1\] is equal to 


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


The value of \[\lim_{x \to \infty} \frac{\left( x + 1 \right)^{10} + \left( x + 2 \right)^{10} + . . . + \left( x + 100 \right)^{10}}{x^{10} + {10}^{10}}\] is 


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


If the value of `lim_(x -> 1) (1 - (1 - x))^"m"/x` is 99, then n = ______.


Evaluate the following limit :

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


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×