Advertisements
Advertisements
प्रश्न
\[\lim_{x \to \frac{\pi}{2}} \frac{\left( \frac{\pi}{2} - x \right) \sin x - 2 \cos x}{\left( \frac{\pi}{2} - x \right) + \cot x}\]
Advertisements
उत्तर
\[\lim_{x \to \frac{\pi}{2}} \frac{\left( \frac{\pi}{2} - x \right) \sin x - 2 \cos x}{\left( \frac{\pi}{2} - x \right) + \cot x}\]
\[ = \lim_{h \to 0} \frac{\left[ \frac{\pi}{2} - \left( \frac{\pi}{2} - h \right) \right] \sin \left( \frac{\pi}{2} - h \right) - 2 \cos \left( \frac{\pi}{2} - h \right)}{\left[ \frac{\pi}{2} - \left( \frac{\pi}{2} - h \right) \right] + \cot \left( \frac{\pi}{2} - h \right)} \left( Plugging x = \frac{\pi}{2} - h \right)\]
\[ = \lim_{h \to 0} \frac{h \cos h - 2 \sin h}{h + \tan h}\]
\[\text{ Dividing the numerator and the denominator by } h:\]
\[ \lim_{h \to 0} \frac{\cos h - 2 \frac{\sin h}{h}}{1 + \frac{\tan h}{h}}\]
\[ \Rightarrow \frac{1 - 2\left( 1 \right)}{1 + 1} = \frac{- 1}{2}\]
APPEARS IN
संबंधित प्रश्न
Show that \[\lim_{x \to 0} \frac{x}{\left| x \right|}\] does not exist.
\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\]
\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]
\[\lim_{x \to 3} \frac{x^2 - 4x + 3}{x^2 - 2x - 3}\]
\[\lim_{x \to 5} \frac{x^2 - 9x + 20}{x^2 - 6x + 5}\]
\[\lim_{x \to \sqrt{3}} \frac{x^2 - 3}{x^2 + 3 \sqrt{3}x - 12}\]
\[\lim_{x \to 1/4} \frac{4x - 1}{2\sqrt{x} - 1}\]
\[\lim_{x \to 1} \frac{x^4 - 3 x^3 + 2}{x^3 - 5 x^2 + 3x + 1}\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = 9,\] find all possible values of a.
\[\lim_{n \to \infty} \left[ \frac{1 + 2 + 3 . . . . . . n - 1}{n^2} \right]\]
\[\lim_{x \to 0} \frac{\tan mx}{\tan nx}\]
\[\lim_{x \to 0} \frac{\sin 5x}{\tan 3x}\]
\[\lim_{x \to 0} \frac{\sin \left( 2 + x \right) - \sin \left( 2 - x \right)}{x}\]
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin \left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\]
\[\lim_\theta \to 0 \frac{1 - \cos 4\theta}{1 - \cos 6\theta}\]
Evaluate the following limits:
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - 1}\]
\[\lim_{x \to a} \frac{\cos x - \cos a}{x - a}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{f\left( x \right) - f\left( \frac{\pi}{4} \right)}{x - \frac{\pi}{4}},\]
\[\lim_{x \to 1} \frac{1 - \frac{1}{x}}{\sin \pi \left( x - 1 \right)}\]
\[\lim_{x \to \frac{3\pi}{2}} \frac{1 + {cosec}^3 x}{\cot^2 x}\]
\[\lim_{x \to 0} \frac{\log \left( 3 + x \right) - \log \left( 3 - x \right)}{x}\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
\[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
\[\lim_{x \to } \frac{1 - \cos 2x}{x} is\]
\[\lim_{x \to 0} \frac{x}{\tan x} is\]
\[\lim_{x \to 0} \frac{\sin x^0}{x}\]
If \[f\left( x \right) = \left\{ \begin{array}{l}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0\end{array}, \right.\] then \[\lim_{x \to 0} f\left( x \right)\] equals
If α is a repeated root of ax2 + bx + c = 0, then \[\lim_{x \to \alpha} \frac{\tan \left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2}\]
The value of \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\]
Evaluate the following limits: `lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`
Evaluate the following Limit:
`lim_(x -> 0) ((1 + x)^"n" - 1)/x`
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->5)[(x^3-125)/(x^5-3125)]`
