Advertisements
Advertisements
Question
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}}\]
Advertisements
Solution
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}}\]
\[ = \lim_{h \to 0} \frac{1 - \tan \left( \frac{\pi}{4} + h \right)}{\left( \frac{\pi}{4} + h - \frac{\pi}{4} \right)}\]
\[ = \lim_{h \to 0} \frac{1 - \left( \frac{1 + \tan h}{1 - \tan h} \right)}{h} \left[ \because \tan \left( \frac{\pi}{4} + 0 \right) = \frac{1 + \tan \theta}{1 - \tan \theta} \right]\]
\[ = \lim_{h \to 0} \frac{1 - \tan h - 1 - \tan h}{\left( 1 - \tan h \right) h}\]
\[ = \lim_{h \to 0} \frac{- 2 \tan h}{h\left( 1 - \tan h \right)} \left[ \because \lim_{h \to 0} \frac{\tan h}{h} = 1 \right]\]
\[ \Rightarrow \frac{- 2}{1 - 0} = - 2\]
APPEARS IN
RELATED QUESTIONS
Show that \[\lim_{x \to 0} \frac{x}{\left| x \right|}\] does not exist.
\[\lim_{x \to 0} 9\]
\[\lim_{x \to - 1}{\left( 4 x^2 + 2 \right)}\]
\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]
\[\lim_{x \to 3} \frac{x^2 - 4x + 3}{x^2 - 2x - 3}\]
\[\lim_{x \to 3} \frac{x^4 - 81}{x^2 - 9}\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{4}{x^3 - 2 x^2} \right)\]
\[\lim_{x \to 1} \frac{\sqrt{x^2 - 1} + \sqrt{x - 1}}{\sqrt{x^2 - 1}}, x > 1\]
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
\[\lim_{x \to \infty} \sqrt{x + 1} - \sqrt{x}\]
\[\lim_{n \to \infty} \frac{n^2}{1 + 2 + 3 + . . . + n}\]
\[\lim_{n \to \infty} \left[ \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!} \right]\]
\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\]
\[f\left( x \right) = \frac{a x^2 + b}{x^2 + 1}, \lim_{x \to 0} f\left( x \right) = 1\] and \[\lim_{x \to \infty} f\left( x \right) = 1,\]then prove that f(−2) = f(2) = 1
Evaluate: \[\lim_{n \to \infty} \frac{1^4 + 2^4 + 3^4 + . . . + n^4}{n^5} - \lim_{n \to \infty} \frac{1^3 + 2^3 + . . . + n^3}{n^5}\]
\[\lim_{x \to 0} \frac{\sin x \cos x}{3x}\]
\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\]
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin \left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to 0} \frac{2 \sin x^\circ - \sin 2 x^\circ}{x^3}\]
\[\lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}\]
Evaluate the following limit:
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{1 - \cos6x}}{\sqrt{2}\left( \frac{\pi}{3} - x \right)}\]
\[\lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{{cosec}^2 x - 2}{\cot x - 1}\]
\[\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n\]
\[\lim_{x \to 0} \left( \cos x + a \sin bx \right)^{1/x}\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to
The value of \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\]
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.
Evaluate the following limit:
`lim_(x->5)[(x^3-125)/(x^5-3125)]`
Evaluate the following limit:
`lim_(x->7)[((root(3)(x)-root(3)(7))(root(3)(x)+root(3)(7)))/(x-7)]`
Evaluate the following limit:
`lim_(x->7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
Evaluate the Following limit:
`lim_ (x -> 3) [sqrt (x + 6)/ x]`
Evaluate the Following limit:
`lim_(x->7)[[(root[3][x] - root[3][7])(root[3][x] + root[3][7])] / (x - 7)]`
