Advertisements
Advertisements
प्रश्न
\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
Advertisements
उत्तर
\[\lim_{x \to \pi} \left( \frac{\sin x}{x - \pi} \right)\]
\[\text{ LHL }: \]
\[ \lim_{x \to \pi^-} \left( \frac{\sin x}{x - \pi} \right)\]
\[\text{ If } x = \pi - h, \text{ then } h \to 0 . \]
\[ = \lim_{h \to 0} \left( \frac{\sin \left( \pi - h \right)}{\pi - h - \pi} \right)\]
\[ = \lim_{h \to 0} \left( \frac{\sin h}{- h} \right)\]
\[ = - 1\]
\[\text{ RHL }: \]
\[ \lim_{x \to \pi^+} \left( \frac{\sin x}{x - \pi} \right)\]
\[If x = \pi + h, \text{ then } h \to 0 . \]
\[ = \lim_{h \to 0} \left( \frac{\sin \left( \pi + h \right)}{\pi + h - \pi} \right)\]
\[ = \lim_{h \to 0} \left( \frac{- \sin h}{h} \right)\]
\[ = - 1\]
\[ \therefore \lim_{x \to \pi} \left( \frac{\sin x}{x - \pi} \right) = - 1\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 2} \left( 3 - x \right)\]
\[\lim_{x \to 0} \frac{3x + 1}{x + 3}\]
\[\lim_{x \to 3} \frac{x^2 - 4x + 3}{x^2 - 2x - 3}\]
\[\lim_{x \to 2} \frac{x^4 - 16}{x - 2}\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to 27} \frac{\left( x^{1/3} + 3 \right) \left( x^{1/3} - 3 \right)}{x - 27}\]
\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\]
\[\lim_{x \to \infty} \frac{5 x^3 - 6}{\sqrt{9 + 4 x^6}}\]
\[\lim_{x \to \infty} \frac{x}{\sqrt{4 x^2 + 1} - 1}\]
`lim_(x->∞) [x{sqrt(x^2+1) - sqrt(x^2-1)}]`
\[\lim_{n \to \infty} \left[ \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + . . . + \frac{1}{3^n} \right]\]
Evaluate: \[\lim_{n \to \infty} \frac{1 . 2 + 2 . 3 + 3 . 4 + . . . + n\left( n + 1 \right)}{n^3}\]
\[\lim_{x \to 0} \frac{\tan mx}{\tan nx}\]
\[\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{2x - \sin x}{\tan x + x}\]
\[\lim_{x \to 0} \frac{\tan 3x - 2x}{3x - \sin^2 x}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x}{\cos 2x - \cos 8x}\]
\[\lim_\theta \to 0 \frac{\sin 4\theta}{\tan 3\theta}\]
Evaluate the following limit:
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin\left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}}\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a}\]
\[\lim_{n \to \infty} 2^{n - 1} \sin \left( \frac{a}{2^n} \right)\]
\[\lim_{x \to 1} \left( 1 - x \right) \tan \left( \frac{\pi x}{2} \right)\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\cos x - \sin x}{\left( \frac{\pi}{4} - x \right) \left( \cos x + \sin x \right)}\]
\[\lim_{x \to \frac{3\pi}{2}} \frac{1 + {cosec}^3 x}{\cot^2 x}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log a}{x}\]
Write the value of \[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]
\[\lim_{x \to 0} \frac{x}{\tan x} is\]
\[\lim_{x \to \infty} \frac{\sqrt{x^2 - 1}}{2x + 1}\]
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\] is equal to
\[\lim_{x \to 0} \frac{8}{x^8}\left\{ 1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cos \frac{x^2}{4} \right\}\] is equal to
The value of \[\lim_{x \to 0} \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} - \sqrt{a - x}}\]
If the value of `lim_(x -> 1) (1 - (1 - x))^"m"/x` is 99, then n = ______.
If `lim_(x -> 1) (x^4 - 1)/(x - 1) = lim_(x -> k) (x^3 - l^3)/(x^2 - k^2)`, then find the value of k.
Evaluate the following limit:
`lim_(x->5)[(x^3-125)/(x^5-3125)]`
Evaluate the following limit:
`\underset{x->3}{lim}[sqrt(x +6)/(x)]`
Evaluate the following limit:
`\underset{x->5}{lim}[(x^3 - 125)/(x^5 - 3125)]`
