Advertisements
Advertisements
प्रश्न
The value of \[\lim_{x \to 0} \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} - \sqrt{a - x}}\]
पर्याय
a
\[\sqrt{a}\]
−a
\[- \sqrt{a}\]
उत्तर
\[- \sqrt{a}\]
\[\lim_{x \to 0} \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} - \sqrt{a - x}}\]
\[\text{ Multiplying } \left( \sqrt{a^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right) and \left( \sqrt{a + x} + \sqrt{a - x} \right) in N^r and D^r : \]
\[ \lim_{x \to 0} \frac{\left( a^2 + x^2 - ax - a^2 - ax - x^2 \right) \left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( a + x - a + x \right) \left( \sqrt{x^2 - ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}\]
\[ \lim_{x \to 0} \frac{- 2ax \left( \sqrt{a + x} + \sqrt{a - x} \right)}{2x \left( \sqrt{a^2 + ax + x^2} + \sqrt{a^2 + ax + x^2} \right)}\]
\[ = \frac{- a\left( 2\sqrt{a} \right)}{2a} = - \sqrt{a}\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\]
\[\lim_{x \to - 1}{\left( 4 x^2 + 2 \right)}\]
\[\lim_{x \to 0} \frac{3x + 1}{x + 3}\]
\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\]
\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\]
\[\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\]
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
\[\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}\]
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . n^3}{\left( n - 1 \right)^4} \right]\]
\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number.
\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\]
\[\lim_{x \to 0} \frac{3 \sin^2 x - 2 \sin x^2}{3 x^2}\]
\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]
\[\lim_{x \to 0} \frac{cosec x - \cot x}{x}\]
\[\lim_{x \to a} \frac{\cos x - \cos a}{x - a}\]
\[\lim_{x \to 1} \frac{1 - \frac{1}{x}}{\sin \pi \left( x - 1 \right)}\]
\[\lim_{n \to \infty} \frac{\sin \left( \frac{a}{2^n} \right)}{\sin \left( \frac{b}{2^n} \right)}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n\]
\[\lim_{x \to 0} \left( \cos x \right)^{1/\sin x}\]
Write the value of \[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
If \[f\left( x \right) = x \sin \left( 1/x \right), x \neq 0,\] then \[\lim_{x \to 0} f\left( x \right) =\]
\[\lim_{n \to \infty} \left\{ \frac{1}{1 - n^2} + \frac{2}{1 - n^2} + . . . + \frac{n}{1 - n^2} \right\}\]
\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to
\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]
\[\lim_{x \to \pi/3} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\] is equal to
\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] is equal to
\[\lim_\theta \to \pi/2 \frac{1 - \sin \theta}{\left( \pi/2 - \theta \right) \cos \theta}\] is equal to
Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
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->7)[[(root3(x)- root3(7))(root3(x) + root3(7)))/(x-7)]`
Evaluate the following limit:
`lim_(x->3)[sqrt(x+6)/x]`
Evaluate the Following limit:
`lim_ (x -> 3) [sqrt (x + 6)/ x]`
