Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x} - \sqrt{3}}}{x - 2}\] is equal to
पर्याय
\[\frac{1}{8\sqrt{3}}\]
\[\frac{1}{\sqrt{3}}\]
$\mathnormal{8 \sqrt{3}}$
\[\sqrt{3}\]
Advertisements
उत्तर
\[(a)\]
\[ \because \lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2} = \lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2} \times \frac{\sqrt{1 + \sqrt{2 + x}} + \sqrt{3}}{\sqrt{1 + \sqrt{2 + x}} + \sqrt{3}}\]
\[ = \lim_{x \to 2} \frac{\sqrt{2 + x} - 2}{\left( x - 2 \right)\left( \sqrt{1 + \sqrt{2 + x}} + \sqrt{3} \right)}\]
\[ = \lim_{x \to 2} \frac{\sqrt{2 + x} - 2}{\left( x - 2 \right)\left( \sqrt{1 + \sqrt{2 + x}} + \sqrt{3} \right)} \times \frac{\sqrt{2 + x} + 2}{\sqrt{2 + x} + 2}\]
\[ = \lim_{x \to 2} \frac{\left( x - 2 \right)}{\left( x - 2 \right)\left( \sqrt{1 + \sqrt{2 + x}} + \sqrt{3} \right)\left( \sqrt{2 + x} + 2 \right)}\]
\[ = \lim_{x \to 2} \frac{1}{\left( \sqrt{1 + \sqrt{2 + x}} + \sqrt{3} \right)\left( \sqrt{2 + x} + 2 \right)}\]
\[ = \frac{1}{\left( \sqrt{1 + \sqrt{2 + 2}} + \sqrt{3} \right)\left( \sqrt{2 + 2} + 2 \right)}\]
\[ = \frac{1}{2\sqrt{3} \times 4}\]
\[ = \frac{1}{8\sqrt{3}}\]
\[\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 1} \frac{x^2 + 1}{x + 1}\]
\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\]
\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\]
\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\]
\[\lim_{x \to 0} \frac{\left( a + x \right)^2 - a^2}{x}\]
\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\]
If \[\lim_{x \to a} \frac{x^3 - a^3}{x - a} = \lim_{x \to 1} \frac{x^4 - 1}{x - 1},\] find all possible values of a.
\[\lim_{x \to \infty} \frac{\sqrt{x^2 + a^2} - \sqrt{x^2 + b^2}}{\sqrt{x^2 + c^2} - \sqrt{x^2 + d^2}}\]
\[\lim_{n \to \infty} \left[ \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + . . . + \frac{1}{3^n} \right]\]
\[\lim_{x \to 0} \frac{\sin x^0}{x}\]
\[\lim_\theta \to 0 \frac{\sin 3\theta}{\tan 2\theta}\]
\[\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x + \tan^2 x}{x \sin x}\]
\[\lim_{x \to 0} \frac{x^2 - \tan 2x}{\tan x}\]
\[\lim_{x \to 0} \frac{2 \sin x^\circ - \sin 2 x^\circ}{x^3}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\sin 2x}{\cos x}\]
\[\lim_{x \to a} \frac{\cos x - \cos a}{x - a}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}}\]
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\sqrt{2 - \sin x} - 1}{\left( \frac{\pi}{2} - x \right)^2}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{f\left( x \right) - f\left( \frac{\pi}{4} \right)}{x - \frac{\pi}{4}},\]
\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
\[\lim_{x \to 1} \left( 1 - x \right) \tan \left( \frac{\pi x}{2} \right)\]
\[\lim_{x \to \frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \tan x\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to \frac{3\pi}{2}} \frac{1 + {cosec}^3 x}{\cot^2 x}\]
\[\lim_{x \to 0^+} \left\{ 1 + \tan^2 \sqrt{x} \right\}^{1/2x}\]
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]
\[\lim_{x \to 0^-} \frac{\sin x}{\sqrt{x}} .\]
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
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
\[\lim_{x \to 1} \left[ x - 1 \right]\] where [.] is the greatest integer function, is equal to
If \[f\left( x \right) = \begin{cases}\frac{\sin\left[ x \right]}{\left[ x \right]}, & \left[ x \right] \neq 0 \\ 0, & \left[ x \right] = 0\end{cases}\] where denotes the greatest integer function, then \[\lim_{x \to 0} f\left( x \right)\]
Evaluate the following limit:
`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
Evaluate the following Limit:
`lim_(x -> 0) ((1 + x)^"n" - 1)/x`
`lim_(x->3) (x^5 - 243)/(x^3 - 27)` = ?
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->3)[sqrt(x+6)/x]`
Evaluate the following limits: `lim_(x -> 3) [sqrt(x + 6)/x]`
