Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 1} \frac{1 - x^2}{\sin 2\pi x}\]
Advertisements
उत्तर
\[\lim_{x \to 1} \frac{1 - x^2}{\sin 2\pi x}\]
\[ = \lim_{h \to 0} \frac{1 - \left( 1 - h \right)^2}{\sin 2\pi\left( 1 - h \right)}\]
\[ = \lim_{h \to 0} \frac{2h - h^2}{- \sin 2\pi h}\]
\[ = \lim_{h \to 0} \frac{- \left( 2 - h \right)}{\frac{\sin 2\pi h}{h}}\]
\[ = \lim_{h \to 0} \frac{\left( h - 2 \right)}{\frac{2\pi \sin 2\pi h}{\left( 2\pi h \right)}}\]
\[ = \frac{0 - 2}{2\pi \times 1}\]
\[ = \frac{- 1}{\pi}\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 1} \frac{1 + \left( x - 1 \right)^2}{1 + x^2}\]
\[\lim_{x \to 0} \frac{x^{2/3} - 9}{x - 27}\]
\[\lim_{x \to \sqrt{3}} \frac{x^2 - 3}{x^2 + 3 \sqrt{3}x - 12}\]
\[\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{4}{x^3 - 2 x^2} \right)\]
Evaluate the following limit:
\[\lim_{x \to 1} \frac{x^7 - 2 x^5 + 1}{x^3 - 3 x^2 + 2}\]
\[\lim_{x \to a} \frac{x^{2/3} - a^{2/3}}{x^{3/4} - a^{3/4}}\]
If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = 9,\] find all possible values of a.
\[\lim_{n \to \infty} \frac{n^2}{1 + 2 + 3 + . . . + n}\]
\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]
`lim_(x->∞) [x{sqrt(x^2+1) - sqrt(x^2-1)}]`
\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number.
Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]
\[\lim_{x \to 0} \frac{5 x \cos x + 3 \sin x}{3 x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{x^2 - \tan 2x}{\tan x}\]
\[\lim_{x \to 0} \frac{x^2 + 1 - \cos x}{x \sin x}\]
\[\lim_{x \to 0} \frac{\sin 2x \left( \cos 3x - \cos x \right)}{x^3}\]
\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\]
\[\lim_{x \to 0} \frac{3 \sin^2 x - 2 \sin x^2}{3 x^2}\]
\[\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}\]
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]
\[\lim_{n \to \infty} \frac{\sin \left( \frac{a}{2^n} \right)}{\sin \left( \frac{b}{2^n} \right)}\]
\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\]
\[\lim_{x \to 0} \frac{8^x - 2^x}{x}\]
\[\lim_{x \to 0} \left( \cos x + a \sin bx \right)^{1/x}\]
Write the value of \[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
\[\lim_{x \to 0^-} \frac{\sin x}{\sqrt{x}} .\]
If \[f\left( x \right) = x \sin \left( 1/x \right), x \neq 0,\] then \[\lim_{x \to 0} f\left( x \right) =\]
\[\lim_{x \to 0} \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]
\[\lim 2_{h \to 0} \left\{ \frac{\sqrt{3} \sin \left( \pi/6 + h \right) - \cos \left( \pi/6 + h \right)}{\sqrt{3} h \left( \sqrt{3} \cos h - \sin h \right)} \right\}\]
\[\lim_{x \to 0} \frac{8}{x^8}\left\{ 1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cos \frac{x^2}{4} \right\}\] is equal to
If α is a repeated root of ax2 + bx + c = 0, then \[\lim_{x \to \alpha} \frac{\tan \left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2}\]
\[\lim_{x \to \infty} \frac{\left| x \right|}{x}\] is equal to
Evaluate the following limit:
`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
Evaluate the following Limit:
`lim_(x -> 0) ((1 + x)^"n" - 1)/x`
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]`
