Advertisements
Advertisements
प्रश्न
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}}\]
Advertisements
उत्तर
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}}\]
\[ = \lim_{h \to 0} \frac{1 - \tan \left( \frac{\pi}{4} + h \right)}{\left( \frac{\pi}{4} + h - \frac{\pi}{4} \right)}\]
\[ = \lim_{h \to 0} \frac{1 - \left( \frac{1 + \tan h}{1 - \tan h} \right)}{h} \left[ \because \tan \left( \frac{\pi}{4} + 0 \right) = \frac{1 + \tan \theta}{1 - \tan \theta} \right]\]
\[ = \lim_{h \to 0} \frac{1 - \tan h - 1 - \tan h}{\left( 1 - \tan h \right) h}\]
\[ = \lim_{h \to 0} \frac{- 2 \tan h}{h\left( 1 - \tan h \right)} \left[ \because \lim_{h \to 0} \frac{\tan h}{h} = 1 \right]\]
\[ \Rightarrow \frac{- 2}{1 - 0} = - 2\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to \sqrt{3}} \frac{x^2 - 3}{x^2 + 3 \sqrt{3}x - 12}\]
\[\lim_{x \to 2} \left( \frac{x}{x - 2} - \frac{4}{x^2 - 2x} \right)\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{2}{x^2 - 2x} \right)\]
\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\]
\[\lim_{n \to \infty} \left[ \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!} \right]\]
Evaluate: \[\lim_{n \to \infty} \frac{1 . 2 + 2 . 3 + 3 . 4 + . . . + n\left( n + 1 \right)}{n^3}\]
\[\lim_{x \to 0} \left[ \frac{x^2}{\sin x^2} \right]\]
\[\lim_{x \to 0} \frac{\sin x^n}{x^n}\]
\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x + \tan^2 x}{x \sin x}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{x^2 + 1 - \cos x}{x \sin x}\]
\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\]
\[\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}\]
Evaluate the following limit:
\[\lim_{x \to 0} \frac{\sin\left( \alpha + \beta \right)x + \sin\left( \alpha - \beta \right)x + \sin2\alpha x}{\cos^2 \beta x - \cos^2 \alpha x}\]
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 1} \left( 1 - x \right) \tan \left( \frac{\pi x}{2} \right)\]
\[\lim_{x \to 0^+} \left\{ 1 + \tan^2 \sqrt{x} \right\}^{1/2x}\]
Write the value of \[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
\[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]
Write the value of \[\lim_{x \to 2} \frac{\left| x - 2 \right|}{x - 2} .\]
\[\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + . . . + n^2}{n^3}\]
\[\lim_{x \to 0} \frac{\sin 2x}{x}\]
\[\lim_{x \to 0} \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]
\[\lim_{n \to \infty} \left\{ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right\}\]is equal to
\[\lim_{x \to 1} \frac{\sin \pi x}{x - 1}\]
\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] is equal to
\[\lim_{x \to \infty} a^x \sin \left( \frac{b}{a^x} \right), a, b > 1\] is equal to
The value of \[\lim_{n \to \infty} \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!}\] is
If \[f\left( x \right) = \begin{cases}\frac{\sin\left[ x \right]}{\left[ x \right]}, & \left[ x \right] \neq 0 \\ 0, & \left[ x \right] = 0\end{cases}\] where denotes the greatest integer function, then \[\lim_{x \to 0} f\left( x \right)\]
Evaluate the following limit:
`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^2 - 25)]`
`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.
