English

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

Advertisements
Advertisements

Question

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

Options

  •  2/3 

  • 4/3 

  • \[- 2\sqrt{3}\] 

  • −4/3

MCQ
Advertisements

Solution

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
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.13 [Page 78]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.13 | Q 14 | Page 78

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


\[\lim_{x \to 0} \frac{x^{2/3} - 9}{x - 27}\]


\[\lim_{x \to 3} \frac{x^4 - 81}{x^2 - 9}\] 


\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\] 


\[\lim_{x \to 5} \frac{x^3 - 125}{x^2 - 7x + 10}\] 


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


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


\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\] 


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


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


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


\[\lim_{x \to 0} \frac{3 \sin x - 4 \sin^3 x}{x}\] 


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


\[\lim_{x \to 0} \frac{5 x \cos x + 3 \sin x}{3 x^2 + \tan x}\] 


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


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


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


\[\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}\] 


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


\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]


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


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


Evaluate the following limits: 

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


If  \[\lim_{x \to 0} kx  cosec x = \lim_{x \to 0} x  cosec kx,\] 


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


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


\[\lim_{x \to \frac{\pi}{4}} \frac{\cos x - \sin x}{\left( \frac{\pi}{4} - x \right) \left( \cos x + \sin x \right)}\]


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


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


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


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


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


\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\] 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 \[\lim_{n \to \infty} \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!}\] 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 limit:

`\underset{x->5}{lim}[(x^3 - 125)/(x^5 - 3125)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×