English

Im X → 0 1 − Cos 4 X X 2 - Mathematics

Advertisements
Advertisements

Question

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

Advertisements

Solution

\[\lim_{x \to 0} \left[ \frac{1 - \cos \left( 4x \right)}{x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \sin^2 2x}{x^2} \right]\]
\[ = \lim_{x \to 0} 2\left[ \frac{\sin 2x}{x} \times \frac{\sin 2x}{x} \right]\]
\[ = \lim_{x \to 0} \left[ 2 \times \frac{\sin 2x}{2x} \times \frac{\sin2x}{2x} \times 4 \right]\]
\[ = 2 \times 1 \times 1 \times 4\]
\[ = 8\]

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 45 | Page 51

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


\[\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\] 


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


\[\lim_{x \to 1} \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1}\]


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


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


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


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


\[\lim_{x \to \infty} \sqrt{x^2 + cx - x}\] 


`lim_(x->∞) [x{sqrt(x^2+1) - sqrt(x^2-1)}]`


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


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


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


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


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


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


\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\] 


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


\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\] 


\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \sin 2x}{1 + \cos 4x}\] 


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


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


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


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


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


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


\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]


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


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


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


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


\[\lim_{x \to \infty} \frac{\left| x \right|}{x}\]  is equal to 


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 -> 5)[(x^3 - 125)/(x^2 - 25)]`


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 limit :

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


Evaluate the following limit:

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


Evaluate the following limit:

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


Evaluate the following limit.

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×