Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 0} \frac{x^2 + 1 - \cos x}{x \sin x}\]
Advertisements
उत्तर
\[\lim_{x \to 0} \left[ \frac{x^2 + 1 - \cos x}{x \sin x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{x^2 + 2 \sin^2 \frac{x}{2}}{x \sin x} \right]\]
\[\text{ Dividing the numerator and the denominator by } x^2 : \]
\[ \lim_{x \to 0} \left[ \frac{1 + \frac{2 \sin^2 \left( \frac{x}{2} \right)}{x^2}}{\frac{\sin x}{x}} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{1 + \frac{2 \sin^2 \left( \frac{x}{2} \right)}{\left( \frac{x}{2} \right)^2 \times 4}}{\frac{\sin x}{x}} \right]\]
\[ = \frac{1 + \frac{2}{4}}{1}\]
\[ = \frac{\frac{3}{2}}{1}\]
\[ = \frac{3}{2}\]
APPEARS IN
संबंधित प्रश्न
Show that \[\lim_{x \to 0} \frac{x}{\left| x \right|}\] does not exist.
\[\lim_{x \to 0} 9\]
\[\lim_{x \to 0} \frac{3x + 1}{x + 3}\]
\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\]
\[\lim_{x \to 2} \left( \frac{x}{x - 2} - \frac{4}{x^2 - 2x} \right)\]
\[\lim_{x \to 1} \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1}\]
\[\lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x^3 - 3 x^2 + 2x} \right]\]
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
\[\lim_{n \to \infty} \left[ \frac{1^2 + 2^2 + . . . + n^2}{n^3} \right]\]
\[\lim_{x \to 0} \frac{\sin 3x}{5x}\]
\[\lim_{x \to 0} \frac{3 \sin x - 4 \sin^3 x}{x}\]
\[\lim_{x \to 0} \frac{7x \cos x - 3 \sin x}{4x + \tan x}\]
\[\lim_{x \to 0} \frac{\sin 5x - \sin 3x}{\sin x}\]
\[\lim_{x \to 0} \frac{\sin \left( a + x \right) + \sin \left( a - x \right) - 2 \sin a}{x \sin x}\]
\[\lim_{x \to 0} \frac{x^2 - \tan 2x}{\tan x}\]
\[\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}\]
\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]
If \[\lim_{x \to 0} kx cosec x = \lim_{x \to 0} x cosec kx,\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\sqrt{2 - \sin x} - 1}{\left( \frac{\pi}{2} - x \right)^2}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin 2\pi x}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{2 - {cosec}^2 x}{1 - \cot x}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
\[\lim_{x \to 0^-} \frac{\sin \left[ x \right]}{\left[ x \right]} .\]
\[\lim_{x \to 0} \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]
\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
If the value of `lim_(x -> 1) (1 - (1 - x))^"m"/x` is 99, then n = ______.
Evaluate `lim_(h -> 0) ((a + h)^2 sin (a + h) - a^2 sina)/h`
`1/(ax^2 + bx + c)`
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.
If `f(x) = {{:(x + 2",", x ≤ - 1),(cx^2",", x > -1):}`, find 'c' if `lim_(x -> -1) f(x)` exists
Evaluate the following limit :
`lim_(x->3)[sqrt(x+6)/x]`
Evaluate the following limit.
`lim_(x->3)[sqrt(x + 6)/x]`
