Advertisements
Advertisements
प्रश्न
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( 4x - \pi \right)^2}\]
Advertisements
उत्तर
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( 4x - \pi \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{2} - \left\{ \cos \left( \frac{\pi}{4} + h \right) + \sin \left( \frac{\pi}{4} + h \right) \right\}}{\left( 4\left( \frac{\pi}{4} + h \right) - \pi \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{2} - \left\{ \cos \frac{\pi}{4} \cos h - \sin \frac{\pi}{4} \sin h + \sin \frac{\pi}{4} \cos h + \cos \frac{\pi}{4} \sin h \right\}}{\left( 4h \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{2} - \sqrt{2}\cos h}{\left( 4h \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{2} \left( 1 - \cos h \right)}{16 h^2}\]
\[ = \lim_{h \to 0} \frac{2\sqrt{2} \sin^2 \frac{h}{2}}{64 \times \frac{h^2}{4}}\]
\[ = \frac{\sqrt{2}}{32} \times \left( 1 \right)^2 \]
\[ = \frac{1}{16\sqrt{2}}\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 2} \left( 3 - x \right)\]
\[\lim_{x \to - 1} \frac{x^3 - 3x + 1}{x - 1}\]
\[\lim_{x \to 3} \frac{x^4 - 81}{x^2 - 9}\]
\[\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\]
\[\lim_{x \to 3} \left( \frac{1}{x - 3} - \frac{3}{x^2 - 3x} \right)\]
\[\lim_{x \to 1} \frac{1 - x^{- 1/3}}{1 - x^{- 2/3}}\]
\[\lim_{x \to 3} \frac{x^2 - x - 6}{x^3 - 3 x^2 + x - 3}\]
Evaluate the following limit:
\[\lim_{x \to 1} \frac{x^7 - 2 x^5 + 1}{x^3 - 3 x^2 + 2}\]
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.
\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\]
Evaluate: \[\lim_{n \to \infty} \frac{1^4 + 2^4 + 3^4 + . . . + n^4}{n^5} - \lim_{n \to \infty} \frac{1^3 + 2^3 + . . . + n^3}{n^5}\]
\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\]
\[\lim_{x \to 0} \frac{\tan^2 3x}{x^2}\]
\[\lim_{x \to 0} \frac{x^2 + 1 - \cos x}{x \sin x}\]
\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\]
\[\lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}\]
\[\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x}{3 \tan^2 x}\]
\[\lim_{x \to 0} \frac{3 \sin x - \sin 3x}{x^3}\]
\[\lim_{x \to 0} \left( cosec x - \cot x \right)\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 1} \frac{1 + \cos \pi x}{\left( 1 - x \right)^2}\]
\[\lim_{n \to \infty} 2^{n - 1} \sin \left( \frac{a}{2^n} \right)\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log a}{x}\]
Write the value of \[\lim_{x \to 0} \frac{\sin x^\circ}{x} .\]
\[\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + . . . + n^2}{n^3}\]
\[\lim_{x \to } \frac{1 - \cos 2x}{x} is\]
\[\lim_{x \to 0} \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]
\[\lim_{x \to \infty} \frac{\sin x}{x}\] equals
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
\[\lim_{x \to \pi/3} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\] is equal to
The value of \[\lim_{x \to \pi/2} \left( \sec x - \tan x \right)\]is
Evaluate the following Limit:
`lim_(x -> 0) ((1 + x)^"n" - 1)/x`
`lim_(x->3) (x^5 - 243)/(x^3 - 27)` = ?
If `f(x) = {{:(x + 2",", x ≤ - 1),(cx^2",", x > -1):}`, find 'c' if `lim_(x -> -1) f(x)` exists
Number of values of x where the function
f(x) = `{{:((tanxlog(x - 2))/(x^2 - 4x + 3); x∈(2, 4) - {3, π}),(1/6tanx; x = 3"," π):}`
is discontinuous, is ______.
Evaluate the Following limit:
`lim_(x->7)[[(root[3][x] - root[3][7])(root[3][x] + root[3][7])] / (x - 7)]`
