English

Lim X → 0 X 2 + 1 − Cos X X Sin X - Mathematics

Advertisements
Advertisements

Question

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

Advertisements

Solution

\[\lim_{x \to 0} \left[ \frac{x^2 + 1 - \cos x}{x \sin x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{x^2 + 2 \sin^2 \frac{x}{2}}{x \sin x} \right]\]
\[\text{ Dividing the numerator and the denominator by } x^2 : \]
\[ \lim_{x \to 0} \left[ \frac{1 + \frac{2 \sin^2 \left( \frac{x}{2} \right)}{x^2}}{\frac{\sin x}{x}} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{1 + \frac{2 \sin^2 \left( \frac{x}{2} \right)}{\left( \frac{x}{2} \right)^2 \times 4}}{\frac{\sin x}{x}} \right]\]
\[ = \frac{1 + \frac{2}{4}}{1}\]
\[ = \frac{\frac{3}{2}}{1}\]
\[ = \frac{3}{2}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: Limits - Exercise 29.7 [Page 50]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 29 Limits
Exercise 29.7 | Q 36 | Page 50

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


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


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


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


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


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


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


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


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


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


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


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


\[\lim_{x \to \frac{\pi}{2}} \frac{\cot x}{\frac{\pi}{2} - x}\]


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


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


\[\lim_{x \to \frac{\pi}{8}} \frac{\cot 4x - \cos 4x}{\left( \pi - 8x \right)^3}\] 


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


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


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


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


\[\lim_{x \to \frac{\pi}{4}} \frac{2 - {cosec}^2 x}{1 - \cot x}\] 


\[\lim_{x \to \frac{3\pi}{2}} \frac{1 + {cosec}^3 x}{\cot^2 x}\]


\[\lim_{x \to 0} \frac{8^x - 2^x}{x}\]


\[\lim_{x \to 0^+} \left\{ 1 + \tan^2 \sqrt{x} \right\}^{1/2x}\]


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


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


Write the value of \[\lim_{x \to \infty} \frac{\sin x}{x} .\] 


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


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


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 \[\lim_{x \to \pi/2} \left( \sec x - \tan x \right)\]is 


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 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)]`


If the value of `lim_(x -> 1) (1 - (1 - x))^"m"/x` is 99, then n = ______.


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 ______.


Evaluate the following limit :

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


Evaluate the following limit:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×