Advertisements
Advertisements
प्रश्न
\[\lim_{x \to \pi/3} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\] is equal to
पर्याय
\[\sqrt{3}\]
- \[\frac{1}{2}\]
\[\frac{1}{\sqrt{3}}\]
\[\sqrt{3}\]
Advertisements
उत्तर
\[\frac{1}{\sqrt{3}}\]
\[\lim_{x \to \frac{\pi}{3}} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\]
\[ = \lim_{h \to 0} \frac{\sin \frac{\pi}{3} - \left( \frac{\pi}{3} - h \right)\left[ \right]}{2 \cos \left( \frac{\pi}{3} - h \right) - 1}\]
\[ = \lim_{h \to 0} \frac{\sin h}{2\left[ \cos \frac{\pi}{3}\cos h + \sin \frac{\pi}{3} \sin h \right] - 1}\]
\[ = \lim_{h \to 0} \frac{\sin h}{2\left[ \frac{1}{2}\cos h + \frac{\sqrt{3}}{2} \sin h \right] - 1}\]
\[ = \lim_{h \to 0} \frac{\sin h}{\cos h + \sqrt{3} \sin h - 1}\]
\[ = \lim_{h \to 0} \frac{\sin h}{- 2 \sin^2 \frac{h}{2} + \sqrt{3} \sin h} \]
\[\text{ Dividing } N^r \text{ and } D^r \text{ by } h: \]
\[ = \lim_{h \to 0} \frac{\frac{\sin h}{h}}{- \left( 2 \times \frac{h}{4} \right) \left( \frac{\sin^2 \frac{h}{2}}{h \times \frac{h}{4}} \right) + \frac{\sqrt{3} \sin h}{h}}\]
\[ = \frac{1}{\sqrt{3}}\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 0} \frac{2 x^2 + 3x + 4}{x^2 + 3x + 2}\]
\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\]
\[\lim_{x \to 1} \left( \frac{1}{x^2 + x - 2} - \frac{x}{x^3 - 1} \right)\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to 3} \left( \frac{1}{x - 3} - \frac{3}{x^2 - 3x} \right)\]
\[\lim_{x \to - 2} \frac{x^3 + x^2 + 4x + 12}{x^3 - 3x + 2}\]
\[\lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x^3 - 3 x^2 + 2x} \right]\]
\[\lim_{x \to a} \frac{x^{5/7} - a^{5/7}}{x^{2/7} - a^{2/7}}\]
If \[\lim_{x \to a} \frac{x^3 - a^3}{x - a} = \lim_{x \to 1} \frac{x^4 - 1}{x - 1},\] find all possible values of a.
\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]
\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\]
\[\lim_{x \to 0} \frac{3 \sin 2x + 2x}{3x + 2 \tan 3x}\]
\[\lim_{x \to 0} \frac{\sin^2 4 x^2}{x^4}\]
\[\lim_{x \to 0} \frac{5 x \cos x + 3 \sin x}{3 x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{\sin 3x - \sin x}{\sin x}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}\]
\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]
Evaluate the following limit:
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin\left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{1 - \sin x}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\left( \frac{\pi}{2} - x \right)^2}\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{x \to \frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \tan x\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\left( \frac{\pi}{2} - x \right) \sin x - 2 \cos x}{\left( \frac{\pi}{2} - x \right) + \cot x}\]
\[\lim_{x \to 0^+} \left\{ 1 + \tan^2 \sqrt{x} \right\}^{1/2x}\]
\[\lim_{x \to 0^-} \frac{\sin \left[ x \right]}{\left[ x \right]} .\]
Write the value of \[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
\[\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + . . . + n^2}{n^3}\]
\[\lim_{x \to 0} \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]
\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to
\[\lim_{x \to \pi/4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}\] is equal to
The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\] is
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: if `lim_(x -> 5)[(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.
if `lim_(x -> 2) (x^"n"- 2^"n")/(x - 2)` = 80 then find the value of n.
Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`
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.
