Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x^3 - 3 x^2 + 2x} \right]\]
Advertisements
उत्तर
\[\lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x^3 - 3 x^2 + 2x} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x\left( x^2 - 3x + 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{- 2\left( 2x - 3 \right)}{x\left( x^2 - 2x - x + 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x\left\{ x\left( x - 2 \right) - 1\left( x - 2 \right) \right\}} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x\left( x - 1 \right)\left( x - 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{x\left( x - 1 \right) - 2\left( 2x - 3 \right)}{x\left( x - 1 \right)\left( x + 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{x^2 - x - 4x + 6}{x\left( x - 1 \right)\left( x + 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{x^2 - 5x + 6}{x\left( x - 1 \right)\left( x + 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{x^2 - 2x - 3x + 6}{x\left( x - 1 \right)\left( x - 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{x\left( x - 2 \right) - 3\left( x - 2 \right)}{x\left( x - 1 \right)\left( x - 2 \right)} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{\left( x - 3 \right)\left( x - 2 \right)}{x\left( x - 1 \right)\left( x - 2 \right)} \right]\]
\[ = \frac{2 - 3}{2\left( 2 - 1 \right)}\]
\[ = - \frac{1}{2}\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\]
\[\lim_{x \to 2} \left( \frac{x}{x - 2} - \frac{4}{x^2 - 2x} \right)\]
\[\lim_{x \to 0} \frac{\left( a + x \right)^2 - a^2}{x}\]
\[\lim_{x \to 3} \left( \frac{1}{x - 3} - \frac{3}{x^2 - 3x} \right)\]
\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\]
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\]
`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.
\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\]
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin \left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to 0} \frac{\tan x - \sin x}{\sin 3x - 3 \sin x}\]
\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x + \tan^2 x}{x \sin x}\]
\[\lim_\theta \to 0 \frac{1 - \cos 4\theta}{1 - \cos 6\theta}\]
Evaluate the following limits:
\[\lim_{x \to 0} \frac{2\sin x - \sin2x}{x^3}\]
\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]
\[\lim_{x \to 0} \left( cosec x - \cot x \right)\]
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{f\left( x \right) - f\left( \frac{\pi}{4} \right)}{x - \frac{\pi}{4}},\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \sin 2x}{1 + \cos 4x}\]
\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
Write the value of \[\lim_{x \to \infty} \frac{\sin x}{x} .\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
\[\lim_{x \to \infty} \frac{\sin x}{x} .\]
If \[f\left( x \right) = x \sin \left( 1/x \right), x \neq 0,\] then \[\lim_{x \to 0} f\left( x \right) =\]
\[\lim_{x \to 0} \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]
\[\lim_{n \to \infty} \left\{ \frac{1}{1 - n^2} + \frac{2}{1 - n^2} + . . . + \frac{n}{1 - n^2} \right\}\]
\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\] is equal at
The value of \[\lim_{x \to 0} \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} - \sqrt{a - x}}\]
The value of \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\]
Evaluate the following limits: if `lim_(x -> 5)[(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.
Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
Evaluate the following Limits: `lim_(x -> "a") ((x + 2)^(5/3) - ("a" + 2)^(5/3))/(x - "a")`
If the value of `lim_(x -> 1) (1 - (1 - x))^"m"/x` is 99, then n = ______.
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)]`
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)]`
