Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x} - \sqrt{3}}}{x - 2}\] is equal to
विकल्प
\[\frac{1}{8\sqrt{3}}\]
\[\frac{1}{\sqrt{3}}\]
$\mathnormal{8 \sqrt{3}}$
\[\sqrt{3}\]
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
Show that \[\lim_{x \to 0} \frac{x}{\left| x \right|}\] does not exist.
\[\lim_{x \to 1} \frac{x^2 + 1}{x + 1}\]
\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]
\[\lim_{x \to - 5} \frac{2 x^2 + 9x - 5}{x + 5}\]
\[\lim_{x \to 3} \frac{x^2 - 4x + 3}{x^2 - 2x - 3}\]
\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\]
\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\]
\[\lim_{x \to \sqrt{3}} \frac{x^2 - 3}{x^2 + 3 \sqrt{3}x - 12}\]
\[\lim_{x \to 2} \left( \frac{x}{x - 2} - \frac{4}{x^2 - 2x} \right)\]
\[\lim_{x \to 1} \frac{1 - x^{- 1/3}}{1 - x^{- 2/3}}\]
\[\lim_{x \to 27} \frac{\left( x^{1/3} + 3 \right) \left( x^{1/3} - 3 \right)}{x - 27}\]
\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\]
\[\lim_{x \to a} \frac{x^{2/3} - a^{2/3}}{x^{3/4} - a^{3/4}}\]
\[\lim_{x \to \infty} \frac{\left( 3x - 1 \right) \left( 4x - 2 \right)}{\left( x + 8 \right) \left( x - 1 \right)}\]
\[\lim_{x \to \infty} \frac{x}{\sqrt{4 x^2 + 1} - 1}\]
\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]
\[\lim_{x \to 0} \frac{\tan mx}{\tan nx}\]
\[\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{5 x \cos x + 3 \sin x}{3 x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^2}\]
\[\lim_{x \to 0} \frac{x^2 + 1 - \cos x}{x \sin x}\]
\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]
\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
\[\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a}\]
\[\lim_{x \to - 1} \frac{x^2 - x - 2}{\left( x^2 + x \right) + \sin \left( x + 1 \right)}\]
\[\lim_{x \to \frac{\pi}{6}} \frac{\cot^2 x - 3}{cosec x - 2}\]
\[\lim_{x \to \frac{3\pi}{2}} \frac{1 + {cosec}^3 x}{\cot^2 x}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log a}{x}\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
If \[f\left( x \right) = x \sin \left( 1/x \right), x \neq 0,\] then \[\lim_{x \to 0} f\left( x \right) =\]
If \[f\left( x \right) = \left\{ \begin{array}{l}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0\end{array}, \right.\] then \[\lim_{x \to 0} f\left( x \right)\] equals
\[\lim_{x \to \infty} a^x \sin \left( \frac{b}{a^x} \right), a, b > 1\] is equal to
\[\lim_{x \to 0} \frac{8}{x^8}\left\{ 1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cos \frac{x^2}{4} \right\}\] is equal to
The value of \[\lim_{x \to 0} \frac{1 - \cos x + 2 \sin x - \sin^3 x - x^2 + 3 x^4}{\tan^3 x - 6 \sin^2 x + x - 5 x^3}\] is
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.
Let `f(x) = {{:((k cos x)/(pi - 2x)",", "when" x ≠ pi/2),(3",", x = pi/2 "and if" f(x) = f(pi/2)):}` find the value of k.
Evaluate the following limit:
`lim_(x->3)[(sqrt(x+6))/x]`
