English

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

Advertisements
Advertisements

Question

Evaluate the following limits: 

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

Advertisements

Solution

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

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

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

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

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

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.7 [Page 51]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.7 | Q 61 | Page 51

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


\[\lim_{x \to - 5} \frac{2 x^2 + 9x - 5}{x + 5}\] 


\[\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\] 


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


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{\tan mx}{\tan nx}\] 


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


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


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


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


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


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


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


\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]


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


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


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


\[\lim_{x \to 0} \frac{8^x - 2^x}{x}\]


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


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


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


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


\[\lim_{x \to 0} \frac{\sin 2x}{x}\] 


If \[f\left( x \right) = x \sin \left( 1/x \right), x \neq 0,\]  then \[\lim_{x \to 0} f\left( x \right) =\] 


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


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


\[\lim_{n \to \infty} \frac{n!}{\left( n + 1 \right)! + n!}\]  is equal to


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


The value of \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\] 


Evaluate the following limit:

`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`


Evaluate the following Limits: `lim_(x -> "a") ((x + 2)^(5/3) - ("a" + 2)^(5/3))/(x - "a")`


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×