हिंदी

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

Advertisements
Advertisements

प्रश्न

Evaluate the following limit: 

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

योग
Advertisements

उत्तर

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

`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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.7 [पृष्ठ ५१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.7 | Q 60 | पृष्ठ ५१

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]


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


Evaluate the following limit:

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


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


\[\lim_{x \to a} \frac{x^{5/7} - a^{5/7}}{x^{2/7} - a^{2/7}}\] 


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


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


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


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


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


\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\] 


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


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


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


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


\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\] 


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


\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]


Evaluate the following limits: 

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


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


\[\lim_{n \to \infty} 2^{n - 1} \sin \left( \frac{a}{2^n} \right)\] 

 


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{x}{\tan x} is\] 


\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\]  is equal at 


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


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


\[\lim_{x \to 1} \left[ x - 1 \right]\] where [.] is the greatest integer function, is equal to 


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 limit:

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


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


Evaluate the following limit :

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


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×