Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 3} \frac{x^2 - 4x + 3}{x^2 - 2x - 3}\]
Advertisements
उत्तर
\[\lim_{x \to 3} \left[ \frac{x^2 - 4x + 3}{x^2 - 2x - 3} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 3} \left[ \frac{x^2 - x - 3x + 3}{x^2 - 3x + x - 3} \right]\]
\[ = \lim_{x \to 3} \left[ \frac{\left( x - 3 \right)\left( x - 1 \right)}{\left( x + 1 \right)\left( x - 3 \right)} \right]\]
\[ = \lim_{x \to 3} \left( \frac{x - 1}{x + 1} \right)\]
\[ = \frac{3 - 1}{3 + 1}\]
\[ = \frac{1}{2}\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 1} \frac{1 + \left( x - 1 \right)^2}{1 + x^2}\]
\[\lim_{x \to 2} \left( 3 - x \right)\]
\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\]
\[\lim_{x \to 5} \frac{x^2 - 9x + 20}{x^2 - 6x + 5}\]
\[\lim_{x \to \sqrt{2}} \frac{x^2 - 2}{x^2 + \sqrt{2}x - 4}\]
\[\lim_{x \to 1} \left( \frac{1}{x - 1} - \frac{2}{x^2 - 1} \right)\]
\[\lim_{x \to 3} \frac{x^2 - x - 6}{x^3 - 3 x^2 + x - 3}\]
\[\lim_{x \to 1} \frac{\sqrt{x^2 - 1} + \sqrt{x - 1}}{\sqrt{x^2 - 1}}, x > 1\]
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\]
If \[\lim_{x \to a} \frac{x^5 - a^5}{x - a} = 405,\]find all possible values of a.
\[\lim_{n \to \infty} \frac{n^2}{1 + 2 + 3 + . . . + n}\]
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . . n^3}{n^4} \right]\]
Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
\[\lim_{x \to 0} \frac{3 \sin 2x + 2x}{3x + 2 \tan 3x}\]
\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\]
\[\lim_{x \to 0} \frac{\sin^2 4 x^2}{x^4}\]
\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]
\[\lim_{x \to 0} \frac{\sin \left( a + x \right) + \sin \left( a - x \right) - 2 \sin a}{x \sin x}\]
\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\]
If \[\lim_{x \to 0} kx cosec x = \lim_{x \to 0} x cosec kx,\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\sqrt{2 - \sin x} - 1}{\left( \frac{\pi}{2} - x \right)^2}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{f\left( x \right) - f\left( \frac{\pi}{4} \right)}{x - \frac{\pi}{4}},\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{x \to \frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \tan x\]
\[\lim_{x \to 0} \frac{\log \left( 3 + x \right) - \log \left( 3 - x \right)}{x}\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
Write the value of \[\lim_{x \to 0} \frac{\sin x^\circ}{x} .\]
\[\lim_{x \to 0^-} \frac{\sin x}{\sqrt{x}} .\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\] is equal to
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 \[\lim_{n \to \infty} \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!}\] is
Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`
`1/(ax^2 + bx + c)`
If `f(x) = {{:(x + 2",", x ≤ - 1),(cx^2",", x > -1):}`, find 'c' if `lim_(x -> -1) f(x)` exists
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)]`
