मराठी

Lim 2 H → 0 ⎧ ⎪ ⎨ ⎪ ⎩ √ 3 Sin ( π / 6 + H ) − Cos ( π / 6 + H ) √ 3 H ( √ 3 Cos H − Sin H ) ⎫ ⎪ ⎬ ⎪ ⎭

Advertisements
Advertisements

प्रश्न

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

पर्याय

  •  2/3 

  • 4/3 

  • \[- 2\sqrt{3}\] 

  • −4/3

MCQ
Advertisements

उत्तर

4/3 

\[\lim_{h \to 0} 2\left[ \frac{\sqrt{3} \sin \left( \pi/6 + h \right) - \cos \left( \pi/6 + h \right)}{\sqrt{3} h\left( \sqrt{3}\cos h - \sin h \right)} \right]\]
\[ = \lim_{h \to 0} 2\frac{\left[ \frac{\sqrt{3}}{2}\cos h + \frac{3}{2} \sin h - \frac{\sqrt{3}}{2}\cos h + \frac{\sin h}{2} \right]}{h\left( 3 \cos h - \sqrt{3} \sin h \right)}\]
\[ = \lim_{h \to 0} 2\left( \frac{2 \sin h}{h} \right) \times \frac{1}{\left( 3 \cos h - \sqrt{3}\sin h \right)}\]
\[ = \lim_{h \to 0} \frac{4}{3 \cos h - \sqrt{3} \sin h}\]
\[ = \frac{4}{3}\]

 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.13 [पृष्ठ ७८]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.13 | Q 14 | पृष्ठ ७८

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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


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


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


\[\lim_{x \to 5} \frac{x^2 - 9x + 20}{x^2 - 6x + 5}\] 


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


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


\[\lim_{x \to \infty} \frac{5 x^3 - 6}{\sqrt{9 + 4 x^6}}\]


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


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


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


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


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


Evaluate the following limits: 

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

 


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


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


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


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

 


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


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


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


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


Write the value of \[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]


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


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


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


\[\lim_{x \to 0} \frac{\sin 2x}{x}\] 


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


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


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


\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to 


If α is a repeated root of ax2 + bx + c = 0, then \[\lim_{x \to \alpha} \frac{\tan \left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2}\]


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 


If \[f\left( x \right) = \begin{cases}\frac{\sin\left[ x \right]}{\left[ x \right]}, & \left[ x \right] \neq 0 \\ 0, & \left[ x \right] = 0\end{cases}\]  where  denotes the greatest integer function, then \[\lim_{x \to 0} f\left( x \right)\]  


Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`


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


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


Evaluate the Following limit:

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


Evaluate the following limit :

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


Evaluate the following limit :

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×