Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 1} \left( \frac{1}{x^2 + x - 2} - \frac{x}{x^3 - 1} \right)\]
Advertisements
उत्तर
\[\lim_{x \to 1} \left[ \frac{\left( x^3 - 1 \right) - x\left( x^2 + x - 2 \right)}{\left( x^2 + x - 2 \right)\left( x^3 - 1 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{\left( x^3 - 1 \right) - x^3 - x^2 + 2x}{\left( x^2 + x - 2 \right)\left( x^3 - 1 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{- x^2 + 2x - 1}{\left( x^2 + x - 2 \right)\left( x - 1 \right)\left( x^2 + x + 1 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{- \left( x^2 - 2x + 1 \right)}{\left( x^2 + x - 2 \right)\left( x - 1 \right)\left( x^2 + x + 1 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{- \left( x - 1 \right)^2}{\left( x^2 + x - 2 \right)\left( x - 1 \right)\left( x^2 + x + 1 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{- \left( x - 1 \right)}{\left( x^2 + 2x - x - 2 \right)\left( x^2 + x + 1 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{- \left( x - 1 \right)}{\left\{ x\left( x + 2 \right) - 1\left( x + 2 \right) \right\}\left( x^2 + x + 1 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{- \left( x - 1 \right)}{\left( x - 1 \right)\left( x + 2 \right)\left( x^2 + x + 1 \right)} \right]\]
\[ = \frac{- 1}{\left( 1 + 2 \right)\left( 1 + 1 + 1 \right)}\]
\[ = \frac{- 1}{9}\]
APPEARS IN
संबंधित प्रश्न
Find `lim_(x -> 5) f(x)`, where f(x) = |x| - 5
Suppose f(x) = `{(a+bx, x < 1),(4, x = 1),(b-ax, x > 1):}` and if `lim_(x -> 1) f(x) = f(1)` what are possible values of a and b?
\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\]
\[\lim_{x \to 2} \left( 3 - x \right)\]
\[\lim_{x \to - 1}{\left( 4 x^2 + 2 \right)}\]
\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]
\[\lim_{x \to 3} \frac{x^4 - 81}{x^2 - 9}\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{4}{x^3 - 2 x^2} \right)\]
\[\lim_{x \to 3} \left( \frac{1}{x - 3} - \frac{3}{x^2 - 3x} \right)\]
\[\lim_{x \to 1} \left( \frac{1}{x - 1} - \frac{2}{x^2 - 1} \right)\]
\[\lim_{x \to 1} \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1}\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
\[\lim_{x \to a} \frac{x^{2/3} - a^{2/3}}{x^{3/4} - a^{3/4}}\]
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
\[\lim_{x \to 0} \frac{\sin x^0}{x}\]
\[\lim_{x \to 0} \frac{x^2}{\sin x^2}\]
\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\]
\[\lim_{x \to 0} \frac{\sin x^n}{x^n}\]
\[\lim_{x \to 0} \frac{\tan^2 3x}{x^2}\]
\[\lim_{x \to 0} \frac{\sin x^2 \left( 1 - \cos x^2 \right)}{x^6}\]
\[\lim_{x \to 0} \frac{2x - \sin x}{\tan x + x}\]
\[\lim_{x \to 0} \frac{\sin \left( a + x \right) + \sin \left( a - x \right) - 2 \sin a}{x \sin x}\]
\[\lim_{x \to 0} \frac{x^2 + 1 - \cos x}{x \sin x}\]
\[\lim_{x \to 0} \frac{\tan 2x - \sin 2x}{x^3}\]
If \[\lim_{x \to 0} kx cosec x = \lim_{x \to 0} x cosec kx,\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\cot x}{\frac{\pi}{2} - x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
\[\lim_{x \to 1} \left( 1 - x \right) \tan \left( \frac{\pi x}{2} \right)\]
Write the value of \[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]
\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
\[\lim_{x \to 0} \frac{\sin x^0}{x}\]
\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to
\[\lim_{n \to \infty} \left\{ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right\}\]is equal to
\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to
Evaluate the following limit:
`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
If `lim_(x -> 1) (x^4 - 1)/(x - 1) = lim_(x -> k) (x^3 - l^3)/(x^2 - k^2)`, then find the value of k.
Evaluate the following limits: `lim_(x -> 3) [sqrt(x + 6)/x]`
