मराठी

Evaluate the following limits: lim x → 0 cos a x − cos b x cos c x − 1 - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate the following limits: 

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

Advertisements

उत्तर

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

\[ = \lim_{x \to 0} \frac{- 2\sin\left( \frac{ax + bx}{2} \right)\sin\left( \frac{ax - bx}{2} \right)}{- 2 \sin^2 \frac{cx}{2}} \left( 1 - \cos2\theta = 2 \sin^2 \theta \right)\]

\[ = \lim_{x \to 0} \frac{\left( \frac{a + b}{2} \right)x \times \frac{\sin\left( \frac{a + b}{2} \right)x}{\left( \frac{a + b}{2} \right)x} \times \left( \frac{a - b}{2} \right)x \times \frac{\sin\left( \frac{a - b}{2} \right)x}{\left( \frac{a - b}{2} \right)x}}{\frac{c^2 x^2}{4} \times \frac{\sin^2 \frac{cx}{2}}{\frac{c^2 x^2}{4}}}\]

\[ = \frac{\left( a + b \right)\left( a - b \right)}{c^2} \times \frac{\lim_{x \to 0} \frac{\sin\left( \frac{a + b}{2} \right)x}{\left( \frac{a + b}{2} \right)x} \times \lim_{x \to 0} \frac{\sin\left( \frac{a - b}{2} \right)x}{\left( \frac{a - b}{2} \right)x}}{\left( \lim_{x \to 0} \frac{\sin\frac{cx}{2}}{\frac{cx}{2}} \right)^2}\]

\[= \frac{a^2 - b^2}{c^2} \times \frac{1 \times 1}{1} \left( \lim_\theta \to 0 \frac{\sin\theta}{\theta} = 1 \right)\]

\[ = \frac{a^2 - b^2}{c^2}\]

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 61 | पृष्ठ ५१

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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


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


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


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


\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]


\[\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}\] 


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


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


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


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


\[f\left( x \right) = \frac{a x^2 + b}{x^2 + 1}, \lim_{x \to 0} f\left( x \right) = 1\] and \[\lim_{x \to \infty} f\left( x \right) = 1,\]then prove that f(−2) = f(2) = 1


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


\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\] 


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


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


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


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


\[\lim_{x \to 0} \frac{1 - \cos 2x}{3 \tan^2 x}\] 


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


Evaluate the following limits: 

\[\lim_{x \to 0} \frac{2\sin x - \sin2x}{x^3}\] 

 


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


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


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


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


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


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


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


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


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


\[\lim 2_{h \to 0} \left\{ \frac{\sqrt{3} \sin \left( \pi/6 + h \right) - \cos \left( \pi/6 + h \right)}{\sqrt{3} h \left( \sqrt{3} \cos h - \sin h \right)} \right\}\] 


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


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


Evaluate the following limits: `lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`


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


Evaluate the following limit:

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


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×