Advertisements
Advertisements
प्रश्न
\[\lim_{x \to \sqrt{2}} \frac{x^2 - 2}{x^2 + \sqrt{2}x - 4}\]
Advertisements
उत्तर
\[\lim_{x \to \sqrt{2}} \left[ \frac{x^2 - 2}{x^2 + \sqrt{2}x - 4} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to \sqrt{2}} \left[ \frac{x^2 - \left( \sqrt{2} \right)^2}{x^2 + 2\sqrt{2}x - \sqrt{2}x - 4} \right]\]
\[ = \lim_{x \to \sqrt{2}} \left[ \frac{\left( x - \sqrt{2} \right)\left( x + \sqrt{2} \right)}{x\left( x + 2\sqrt{2} \right) - \sqrt{2}\left( x + 2\sqrt{2} \right)} \right]\]
\[ = \lim_{x \to \sqrt{2}} \left[ \frac{\left( x - \sqrt{2} \right)\left( x + \sqrt{2} \right)}{\left( x - \sqrt{2} \right)\left( x + 2\sqrt{2} \right)} \right]\]
\[ = \frac{\sqrt{2} + \sqrt{2}}{\sqrt{2} + 2\sqrt{2}}\]
\[ = \frac{2\sqrt{2}}{3\sqrt{2}}\]
\[ = \frac{2}{3}\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 1} \frac{x^2 + 1}{x + 1}\]
\[\lim_{x \to 0} \frac{2 x^2 + 3x + 4}{x^2 + 3x + 2}\]
\[\lim_{x \to 1} \frac{1 + \left( x - 1 \right)^2}{1 + x^2}\]
\[\lim_{x \to 2} \left( 3 - x \right)\]
\[\lim_{x \to - 1}{\left( 4 x^2 + 2 \right)}\]
\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to 3} \frac{x^2 - x - 6}{x^3 - 3 x^2 + x - 3}\]
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\]
\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\]
\[f\left( x \right) = \frac{a x^2 + b}{x^2 + 1}, \lim_{x \to 0} f\left( x \right) = 1\] and \[\lim_{x \to \infty} f\left( x \right) = 1,\]then prove that f(−2) = f(2) = 1
Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]
\[\lim_{x \to 0} \left[ \frac{x^2}{\sin x^2} \right]\]
\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\]
\[\lim_{x \to 0} \frac{\sin^2 4 x^2}{x^4}\]
\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]
\[\lim_{x \to 0} \frac{\sin 2x \left( \cos 3x - \cos x \right)}{x^3}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos 2x}\]
\[\lim_{x \to 0} \frac{cosec x - \cot x}{x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{{cosec}^2 x - 2}{\cot x - 1}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n\]
\[\lim_{x \to 0} \left( \cos x + \sin x \right)^{1/x}\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
Write the value of \[\lim_{x \to 0^-} \frac{\sin \left[ x \right]}{\left[ x \right]} .\]
Write the value of \[\lim_{x \to 2} \frac{\left| x - 2 \right|}{x - 2} .\]
Write the value of \[\lim_{x \to 0} \frac{\sin x^\circ}{x} .\]
\[\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + . . . + n^2}{n^3}\]
\[\lim_{x \to } \frac{1 - \cos 2x}{x} is\]
\[\lim_{x \to 3} \frac{\sum^n_{r = 1} x^r - \sum^n_{r = 1} 3^r}{x - 3}\]is real to
\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] is equal to
Evaluate the following limits: `lim_(x -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`
Evaluate the following Limits: `lim_(x -> "a") ((x + 2)^(5/3) - ("a" + 2)^(5/3))/(x - "a")`
Evaluate the following Limit:
`lim_(x -> 0) ((1 + x)^"n" - 1)/x`
Evaluate the following limit :
`lim_(x->3)[sqrt(x+6)/x]`
