English

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

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

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

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Find `lim_(x -> 5) f(x)`, where f(x)  = |x| - 5


\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\] 


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


\[\lim_{x \to 1/4} \frac{4x - 1}{2\sqrt{x} - 1}\] 


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


\[\lim_{x \to 1} \frac{\sqrt{x^2 - 1} + \sqrt{x - 1}}{\sqrt{x^2 - 1}}, x > 1\] 


\[\lim_{x \to a} \frac{\left( x + 2 \right)^{5/2} - \left( a + 2 \right)^{5/2}}{x - a}\] 


\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\] 


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


\[\lim_{x \to \infty} \frac{\left( 3x - 1 \right) \left( 4x - 2 \right)}{\left( x + 8 \right) \left( x - 1 \right)}\] 


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


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


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


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


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


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


\[\lim_{x \to 0} \frac{1 - \cos 2x}{3 \tan^2 x}\] 


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


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


Evaluate the following limit:

\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{1 - \cos6x}}{\sqrt{2}\left( \frac{\pi}{3} - x \right)}\]


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


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


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


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


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


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


If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal


The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\]  is


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


The value of \[\lim_{x \to 0} \frac{1 - \cos x + 2 \sin x - \sin^3 x - x^2 + 3 x^4}{\tan^3 x - 6 \sin^2 x + x - 5 x^3}\] is 


Evaluate the following limits: `lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`


Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`


Number of values of x where the function

f(x) = `{{:((tanxlog(x - 2))/(x^2 - 4x + 3); x∈(2, 4) - {3, π}),(1/6tanx; x = 3","  π):}`

is discontinuous, is ______.


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


Evaluate the following limits: `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×