Advertisements
Advertisements
प्रश्न
\[\lim_{x \to \infty} \sqrt{x + 1} - \sqrt{x}\]
Advertisements
उत्तर
\[\lim_{x \to \infty} \left( \sqrt{x + 1} - \sqrt{x} \right)\]
It is of the form ∞–∞.
On rationalising, we get:
\[\lim_{x \to \infty} \left( \sqrt{x + 1} - \sqrt{x} \right) \times \left( \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}} \right)\]
\[ = \lim_{x \to \infty} \left( \frac{x + 1 - x}{\left( \sqrt{x + 1} + \sqrt{x} \right)} \right)\]
\[ = \frac{1}{\infty}\]
\[ = 0\]
APPEARS IN
संबंधित प्रश्न
Find `lim_(x -> 5) f(x)`, where f(x) = |x| - 5
\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\]
\[\lim_{x \to 3} \frac{x^2 - 4x + 3}{x^2 - 2x - 3}\]
\[\lim_{x \to 5} \frac{x^3 - 125}{x^2 - 7x + 10}\]
\[\lim_{x \to 1} \left( \frac{1}{x^2 + x - 2} - \frac{x}{x^3 - 1} \right)\]
\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\]
\[\lim_{x \to 1} \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1}\]
\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number.
\[f\left( x \right) = \frac{a x^2 + b}{x^2 + 1}, \lim_{x \to 0} f\left( x \right) = 1\] and \[\lim_{x \to \infty} f\left( x \right) = 1,\]then prove that f(−2) = f(2) = 1
\[\lim_{x \to 0} \frac{\sin 3x}{5x}\]
\[\lim_{x \to 0} \frac{3 \sin x - 4 \sin^3 x}{x}\]
\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\]
\[\lim_{x \to 0} \frac{\tan mx}{\tan nx}\]
\[\lim_{x \to 0} \frac{\sin^2 4 x^2}{x^4}\]
\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x}{\cos 2x - \cos 8x}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x + \tan^2 x}{x \sin x}\]
\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{1 - \sin x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\left( \frac{\pi}{2} - x \right)^2}\]
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{f\left( x \right) - f\left( \frac{\pi}{4} \right)}{x - \frac{\pi}{4}},\]
\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
If \[f\left( x \right) = x \sin \left( 1/x \right), x \neq 0,\] then \[\lim_{x \to 0} f\left( x \right) =\]
\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\] is equal at
\[\lim_{x \to 1} \frac{\sin \pi x}{x - 1}\]
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to
\[\lim_\theta \to \pi/2 \frac{1 - \sin \theta}{\left( \pi/2 - \theta \right) \cos \theta}\] is equal to
The value of \[\lim_{x \to \pi/2} \left( \sec x - \tan x \right)\]is
\[\lim_{x \to 0} \frac{\left| \sin x \right|}{x}\]
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)[(root(3)(1 + x) - sqrt(1 + x))/x]`
Evaluate the following limits: `lim_(x -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`
Evaluate the following Limit:
`lim_(x -> 0) ((1 + x)^"n" - 1)/x`
Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`
Evaluate the following limit :
`lim_(x->3)[sqrt(x+6)/x]`
Evaluate the following limit :
`lim_(x->5)[(x^3-125)/(x^5-3125)]`
Evaluate the following limit:
`\underset{x->3}{lim}[sqrt(x +6)/(x)]`
