मराठी

Write the Value of Lim X → ∞ Sin X X .

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

\[\lim_{x \to \infty} \left( \frac{\sin x}{x} \right)\]
\[\text{ Let } x = \frac{1}{y}\]
\[\text{ If } x \to \infty , \text{ then } y \to 0 . \]
\[ = \lim_{y \to 0} y \cdot \sin \left( \frac{1}{y} \right)\]
\[\text{ LHL }: \]
\[\text{ Let } y = 0 - h\]
\[\text{ If } y \to 0, \text{ then } h \to 0 . \]
\[ = \lim_{h \to 0} \left( \left( 0 - h \right) \times \sin \left( \frac{1}{0 - h} \right) \right)\]
\[0 \text{ times The oscillating number between } -1 \text{ and } 1\]
\[ = 0\]
\[\text{ RHL }: \]
\[ \lim_{y \to 0^+} \left( y \cdot \sin \left( \frac{1}{y} \right) \right)\]
\[Let y = 0 + h\]
\[\text{ If } y \to 0, \text{ then } h \to 0 . \]
\[ = \lim_{h \to 0} h \times \sin \left( \frac{1}{h} \right)\]
\[ = 0 \text{ times The oscillating number between } -1 \text{ and } 1\]
\[ = 0\]

 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: Limits - Exercise 29.12 [पृष्ठ ७७]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.12 | Q 7 | पृष्ठ ७७

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Show that \[\lim_{x \to 0} \frac{x}{\left| x \right|}\] does not exist.


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


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


\[\lim_{x \to - 1}{\left( 4 x^2 + 2 \right)}\]


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


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


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


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


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


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


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


\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]


\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\] 


\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\] 


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


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


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


\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]


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


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


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


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


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


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


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


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


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


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


If \[f\left( x \right) = \left\{ \begin{array}{l}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0\end{array}, \right.\] then \[\lim_{x \to 0} f\left( x \right)\]  equals 


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 


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


\[\lim_{x \to 1} \left[ x - 1 \right]\] where [.] is the greatest integer function, is equal to 


Evaluate the following limit:

`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`


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


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


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->7)[((root(3)(x)-root(3)(7))(root(3)(x)+root(3)(7)))/(x-7)]`


Evaluate the Following limit: 

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


Evaluate the following limit:

`\underset{x->3}{lim}[sqrt(x +6)/(x)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×