English

Evaluate the following limit: limx→0sin⁡(α+β)x+sin⁡(α−β)x+sin⁡2αxcos2⁡βx−cos2⁡αx - Mathematics

Advertisements
Advertisements

Question

Evaluate the following limit: 

\[\lim_{x \to 0} \frac{\sin\left( \alpha + \beta \right)x + \sin\left( \alpha - \beta \right)x + \sin2\alpha x}{\cos^2 \beta x - \cos^2 \alpha x}\]

Sum
Advertisements

Solution

\[\lim_{x \to 0} \frac{\sin\left( \alpha + \beta \right)x + \sin\left( \alpha - \beta \right)x + \sin2\alpha x}{\cos^2 \beta x - \cos^2 \alpha x}\]

`therefore cos^2 beta x - cos^2alpha x`

`= cos^2betax - cos^2alphax + cos^2betax * cos^2alphax - cos^2betax * cos^2alphax`

= `cos^2 beta x - cos^2 beta x * cos^2 alpha x - cos^2 alpha x + cos^2 beta x * cos^2 alpha x`

= `cos^2 beta x (1 - cos^2 alpha x) - cos^2 alpha x (1 - cos^2 beta x)`

= `cos^2 beta x * sin^2 alpha x - cos^2 alpha x * sin^2 beta x`

= `(cos beta x * sin alpha x)^2 - (cos alpha x * sin beta x)`

= `(cos beta x * sin alpha x + cos alpha x * sin beta x)(cos beta x * sin alpha x - cos alpha x * sin beta x)`

= `sin (alpha x + beta x)  sin (alphax - betax)`    ...[sin (A ± B) = sin A cos B ± cos A sin B]

= `sin (alpha + beta)x  sin (alpha - beta)x`

`therefore lim_(x->0) (x{sin (alpha + beta )x + sin (alpha - beta)x + sin 2 alpha x})/(sin (alpha + beta)x  sin (alpha - beta) x)`

`lim_(x->0) ((x{sin (alpha + beta )x + sin (alpha - beta)x + sin 2 alpha x})/x^2)/((sin (alpha + beta)x  sin (alpha - beta) x)/x^2)`

`lim_(x->0) ((sin (alpha + beta)x)/x + (sin (alpha - beta)x)/x + (sin 2 alpha x)/x)/((sin (alpha + beta)x)/x xx (sin (alpha - beta)x)/x)`

`lim_(x->0) ((sin (alpha + beta)x)/((alpha + beta) x) xx (alpha + beta) + (sin (alpha - beta)x)/((alpha - beta)x) xx (alpha - beta) + (sin 2 alpha x)/(2alphax) xx 2alpha)/((sin (alpha + beta)x)/((alpha + beta)x) xx (alpha + beta) xx (sin (alpha - beta)x)/((alpha - beta)x) xx (alpha - beta))`

`= (alpha + beta + alpha - beta + 2alpha)/((alpha + beta)(alpha - beta))`

`= (4alpha)/(alpha^2 - beta^2)`

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

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.7 | Q 60 | Page 51

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

\[\lim_{x \to 0} \frac{2 x^2 + 3x + 4}{x^2 + 3x + 2}\] 


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


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


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


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


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


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


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


If \[\lim_{x \to a} \frac{x^3 - a^3}{x - a} = \lim_{x \to 1} \frac{x^4 - 1}{x - 1},\] find all possible values of a


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


\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number. 


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


\[\lim_{x \to \infty} \frac{\sin x}{x} .\] 


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


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


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


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


The value of \[\lim_{n \to \infty} \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!}\] is 


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_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`


Evaluate the following limit:

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


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)[(root(3)(1 + x) - sqrt(1 + x))/x]`


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


Evaluate the following limit :

`lim_(x->7)[[(root3(x)- root3(7))(root3(x) + root3(7)))/(x-7)]`


Evaluate the following limit:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×