Advertisements
Advertisements
प्रश्न
If \[\lim_{x \to 0} kx cosec x = \lim_{x \to 0} x cosec kx,\]
Advertisements
उत्तर
\[\lim_{x \to 0} kx . cosec x = \lim_{x \to 0} x cosec kx\]
\[ \Rightarrow \lim_{x \to 0} \left[ \frac{kx}{\sin x} \right] = \lim_{x \to 0} \left[ \frac{x}{\sin \left( kx \right)} \right]\]
\[ \Rightarrow k \lim_{x \to 0} \left[ \frac{x}{\sin x} \right] = \lim_{x \to 0} \left[ \frac{kx}{\sin \left( kx \right)} \right] \times \frac{1}{k}\]
\[ \Rightarrow k = \frac{1}{k}\]
\[ \Rightarrow k^2 = 1\]
\[ \Rightarrow k = \pm 1\]
APPEARS IN
संबंधित प्रश्न
Show that \[\lim_{x \to 0} \frac{x}{\left| x \right|}\] does not exist.
\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\]
\[\lim_{x \to 2} \left( 3 - x \right)\]
\[\lim_{x \to - 1}{\left( 4 x^2 + 2 \right)}\]
\[\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\]
\[\lim_{x \to \sqrt{2}} \frac{x^2 - 2}{x^2 + \sqrt{2}x - 4}\]
\[\lim_{x \to \sqrt{3}} \frac{x^2 - 3}{x^2 + 3 \sqrt{3}x - 12}\]
\[\lim_{x \to 1} \left( \frac{1}{x^2 + x - 2} - \frac{x}{x^3 - 1} \right)\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{2}{x^2 - 2x} \right)\]
\[\lim_{x \to 1} \frac{x^4 - 3 x^3 + 2}{x^3 - 5 x^2 + 3x + 1}\]
\[\lim_{x \to 1} \frac{1 - x^{- 1/3}}{1 - x^{- 2/3}}\]
Evaluate the following limit:
\[\lim_{x \to 1} \frac{x^7 - 2 x^5 + 1}{x^3 - 3 x^2 + 2}\]
\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\]
If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = \lim_{x \to 5} \left( 4 + x \right),\] find all possible values of a.
\[\lim_{x \to \infty} \sqrt{x^2 + cx - x}\]
\[\lim_{n \to \infty} \left[ \frac{1 + 2 + 3 . . . . . . n - 1}{n^2} \right]\]
\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\]
\[f\left( x \right) = \frac{a x^2 + b}{x^2 + 1}, \lim_{x \to 0} f\left( x \right) = 1\] and \[\lim_{x \to \infty} f\left( x \right) = 1,\]then prove that f(−2) = f(2) = 1
\[\lim_{x \to 0} \frac{2x - \sin x}{\tan x + x}\]
\[\lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}\]
\[\lim_\theta \to 0 \frac{1 - \cos 4\theta}{1 - \cos 6\theta}\]
\[\lim_\theta \to 0 \frac{\sin 4\theta}{\tan 3\theta}\]
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
\[\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a}\]
\[\lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a}\]
\[\lim_{x \to 1} \frac{1 - \frac{1}{x}}{\sin \pi \left( x - 1 \right)}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( 4x - \pi \right)^2}\]
\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\]
\[\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n\]
\[\lim_{x \to 0} \left( \cos x + \sin x \right)^{1/x}\]
\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
\[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]
Write the value of \[\lim_{x \to 2} \frac{\left| x - 2 \right|}{x - 2} .\]
\[\lim_{x \to 0} \frac{\sin 2x}{x}\]
\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to
If α is a repeated root of ax2 + bx + c = 0, then \[\lim_{x \to \alpha} \frac{\tan \left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2}\]
\[\lim_\theta \to \pi/2 \frac{1 - \sin \theta}{\left( \pi/2 - \theta \right) \cos \theta}\] is equal to
The value of \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\]
