हिंदी

If lim x → 0 k x c o s e c x = lim x → 0 x c o s e c k x , - Mathematics

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

\[\lim_{x \to 0} kx . cosec x = \lim_{x \to 0} x cosec kx\]
\[ \Rightarrow \lim_{x \to 0} \left[ \frac{kx}{\sin x} \right] = \lim_{x \to 0} \left[ \frac{x}{\sin \left( kx \right)} \right]\]
\[ \Rightarrow k \lim_{x \to 0} \left[ \frac{x}{\sin x} \right] = \lim_{x \to 0} \left[ \frac{kx}{\sin \left( kx \right)} \right] \times \frac{1}{k}\]
\[ \Rightarrow k = \frac{1}{k}\]
\[ \Rightarrow k^2 = 1\]
\[ \Rightarrow k = \pm 1\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: Limits - Exercise 29.7 [पृष्ठ ५१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.7 | Q 63 | पृष्ठ ५१

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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


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


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


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


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


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


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


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


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


\[\lim_{x \to \infty} \frac{\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2}}{\sqrt{x^2 + c^2} - \sqrt{x^2 + d^2}}\] 


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


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


\[\lim_{x \to 0} \frac{\sin 5x}{\tan 3x}\] 


\[\lim_{x \to 0} \frac{7x \cos x - 3 \sin x}{4x + \tan x}\] 


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


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


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


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


\[\lim_{x \to a} \frac{\cos x - \cos a}{x - a}\] 


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


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


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


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


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


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


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


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


\[\lim_{x \to \pi/3} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\] 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 \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\] 


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 


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


Evaluate the following limits: `lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`


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)]`


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


Let `f(x) = {{:((k cos x)/(pi - 2x)",", "when"  x ≠ pi/2),(3",", x = pi/2  "and if"  f(x) = f(pi/2)):}` find the value of k.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×