Advertisements
Advertisements
प्रश्न
\[\lim_{x \to a} \frac{\cos x - \cos a}{x - a}\]
Advertisements
उत्तर
\[ = \lim_{x \to a} \frac{- 2 \sin \left( \frac{x + a}{2} \right) \sin \left( \frac{x - a}{2} \right)}{2\left( \frac{x - a}{2} \right)} \left[ \because \cos A - \cos B - 2 \sin \left( \frac{A - B}{2} \right) \sin \left( \frac{A + B}{2} \right) \right]\]
\[ = \lim_{x \to a} - \sin \left( \frac{x + a}{2} \right) \left[ \because \lim_\theta \to a \sin\frac{\left( \theta - a \right)}{\left( \theta - a \right)} = 1 \right]\]
\[ = - \sin \left( \frac{2a}{2} \right)\]
\[ \Rightarrow - \sin a\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 1} \frac{1 + \left( x - 1 \right)^2}{1 + x^2}\]
\[\lim_{x \to 0} 9\]
\[\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\]
\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\]
\[\lim_{x \to 1} \left( \frac{1}{x^2 + x - 2} - \frac{x}{x^3 - 1} \right)\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to 1/4} \frac{4x - 1}{2\sqrt{x} - 1}\]
\[\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}\]
If \[\lim_{x \to a} \frac{x^5 - a^5}{x - a} = 405,\]find all possible values of a.
\[\lim_{x \to \infty} \frac{\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2}}{\sqrt{x^2 + c^2} - \sqrt{x^2 + d^2}}\]
\[\lim_{n \to \infty} \left[ \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!} \right]\]
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin \left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to 0} \frac{\tan x - \sin x}{\sin 3x - 3 \sin x}\]
\[\lim_{x \to 0} \frac{x \cos x + \sin x}{x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{\tan 2x - \sin 2x}{x^3}\]
\[\lim_{x \to 1} \frac{1 - \frac{1}{x}}{\sin \pi \left( x - 1 \right)}\]
\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
\[\lim_{x \to 1} \left( 1 - x \right) \tan \left( \frac{\pi x}{2} \right)\]
\[\lim_{x \to \frac{\pi}{4}} \frac{{cosec}^2 x - 2}{\cot x - 1}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 0} \left( \cos x + \sin x \right)^{1/x}\]
Write the value of \[\lim_{x \to 0} \frac{\sin x^\circ}{x} .\]
\[\lim_{x \to \infty} \frac{\sin x}{x}\] equals
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to
\[\lim_{x \to 0} \frac{8}{x^8}\left\{ 1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cos \frac{x^2}{4} \right\}\] is equal to
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: `lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`
`1/(ax^2 + bx + c)`
If f(x) = `{{:(1 if x "is rational"),(-1 if x "is rational"):}` is continuous on ______.
