Advertisements
Advertisements
प्रश्न
\[\lim 2_{h \to 0} \left\{ \frac{\sqrt{3} \sin \left( \pi/6 + h \right) - \cos \left( \pi/6 + h \right)}{\sqrt{3} h \left( \sqrt{3} \cos h - \sin h \right)} \right\}\]
पर्याय
2/3
4/3
\[- 2\sqrt{3}\]
−4/3
Advertisements
उत्तर
4/3
\[\lim_{h \to 0} 2\left[ \frac{\sqrt{3} \sin \left( \pi/6 + h \right) - \cos \left( \pi/6 + h \right)}{\sqrt{3} h\left( \sqrt{3}\cos h - \sin h \right)} \right]\]
\[ = \lim_{h \to 0} 2\frac{\left[ \frac{\sqrt{3}}{2}\cos h + \frac{3}{2} \sin h - \frac{\sqrt{3}}{2}\cos h + \frac{\sin h}{2} \right]}{h\left( 3 \cos h - \sqrt{3} \sin h \right)}\]
\[ = \lim_{h \to 0} 2\left( \frac{2 \sin h}{h} \right) \times \frac{1}{\left( 3 \cos h - \sqrt{3}\sin h \right)}\]
\[ = \lim_{h \to 0} \frac{4}{3 \cos h - \sqrt{3} \sin h}\]
\[ = \frac{4}{3}\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to - 1} \frac{x^3 - 3x + 1}{x - 1}\]
\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\]
\[\lim_{x \to 1} \frac{x^4 - 3 x^3 + 2}{x^3 - 5 x^2 + 3x + 1}\]
\[\lim_{x \to a} \frac{\left( x + 2 \right)^{5/2} - \left( a + 2 \right)^{5/2}}{x - a}\]
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
\[\lim_{x \to a} \frac{x^{2/7} - a^{2/7}}{x - a}\]
\[\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}\]
\[\lim_{x \to \infty} \sqrt{x^2 + cx - x}\]
\[\lim_{x \to \infty} \sqrt{x + 1} - \sqrt{x}\]
\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . . n^3}{n^4} \right]\]
\[\lim_{x \to 0} \frac{3 \sin x - 4 \sin^3 x}{x}\]
\[\lim_{x \to 0} \frac{\sin^2 4 x^2}{x^4}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}\]
Evaluate the following limit:
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{1 - \cos6x}}{\sqrt{2}\left( \frac{\pi}{3} - x \right)}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \sin 2x}{1 + \cos 4x}\]
\[\lim_{x \to \frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \tan x\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( 4x - \pi \right)^2}\]
Evaluate the following limit:
\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]
\[\lim_{x \to 0} \left( \cos x + \sin x \right)^{1/x}\]
Write the value of \[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]
\[\lim_{x \to 0^-} \frac{\sin \left[ x \right]}{\left[ x \right]} .\]
\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
\[\lim_{x \to 0} \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]
\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to
\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to
The value of \[\lim_{x \to \pi/2} \left( \sec x - \tan x \right)\]is
If \[f\left( x \right) = \begin{cases}\frac{\sin\left[ x \right]}{\left[ x \right]}, & \left[ x \right] \neq 0 \\ 0, & \left[ x \right] = 0\end{cases}\] where denotes the greatest integer function, then \[\lim_{x \to 0} f\left( x \right)\]
Evaluate the following limits: `lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`
Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^2 - 25)]`
`lim_(x->3) (x^5 - 243)/(x^3 - 27)` = ?
Which of the following function is not continuous at x = 0?
Evaluate `lim_(h -> 0) ((a + h)^2 sin (a + h) - a^2 sina)/h`
Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`
Evaluate the following limit.
`lim_(x->3)[sqrt(x + 6)/x]`
