Advertisements
Advertisements
प्रश्न
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]
Advertisements
उत्तर
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x} \]
\[ = \lim_{x \to \infty} \left\{ 1 + \frac{3 x^2 + 1}{4 x^2 - 1} - 1 \right\}^\frac{x^3}{1 + x} \]
\[ = \lim_{x \to \infty} \left\{ 1 + \frac{3 x^2 + 1 - 4 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x} \]
\[ = \lim_{x \to \infty} \left\{ 1 + \frac{2 - x^2}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x} \]
\[ = e^\lim_{x \to \infty} \left( \frac{2 - x^2}{4 x^2 - 1} \right) \times \left( \frac{x^3}{1 + x} \right) \]
\[ = e {}^\lim_{x \to \infty} \left( \frac{- x^5 + 2 x^3}{\left( 4 x^2 - 1 \right) \left( 1 + x \right)} \right) \]
\[\text{ Dividing N^r and } D^r \text{ by } x^5 : \]
\[ = e {}^\lim_{x \to \infty} \left( \frac{- 1 + \frac{2}{x^3}}{\frac{\left( 4 x^2 - 1 \right)}{x^3} \frac{\left( 1 + x \right)}{x^2}} \right) \]
\[ = e {}^\lim_{x \to \infty} \left( \frac{- 1 + \frac{2}{x^3}}{\left( \frac{4}{x} - \frac{1}{x^3} \right) \left( \frac{1}{x^2} + \frac{1}{x} \right)} \right) \]
\[ = e^{- \frac{1}{0}} \]
\[ = e^{- \infty} = 0\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 0} 9\]
\[\lim_{x \to 2} \frac{x^4 - 16}{x - 2}\]
\[\lim_{x \to \sqrt{3}} \frac{x^2 - 3}{x^2 + 3 \sqrt{3}x - 12}\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to - 2} \frac{x^3 + x^2 + 4x + 12}{x^3 - 3x + 2}\]
\[\lim_{x \to 1} \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1}\]
\[\lim_{x \to 1} \frac{\sqrt{x^2 - 1} + \sqrt{x - 1}}{\sqrt{x^2 - 1}}, x > 1\]
\[\lim_{x \to a} \frac{\left( x + 2 \right)^{5/2} - \left( a + 2 \right)^{5/2}}{x - a}\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
\[\lim_{x \to 0} \frac{1 - \cos mx}{x^2}\]
\[\lim_\theta \to 0 \frac{\sin 3\theta}{\tan 2\theta}\]
\[\lim_{x \to 0} \frac{\sin 3x - \sin x}{\sin x}\]
\[\lim_{x \to 0} \frac{\tan 3x - 2x}{3x - \sin^2 x}\]
\[\lim_{x \to 0} \frac{\sin \left( 2 + x \right) - \sin \left( 2 - x \right)}{x}\]
\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]
\[\lim_{x \to 0} \frac{3 \sin^2 x - 2 \sin x^2}{3 x^2}\]
\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]
\[\lim_{x \to 0} \left( cosec x - \cot x \right)\]
Evaluate the following limit:
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin\left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]
\[\lim_{x \to 1} \frac{1 + \cos \pi x}{\left( 1 - x \right)^2}\]
\[\lim_{n \to \infty} \frac{\sin \left( \frac{a}{2^n} \right)}{\sin \left( \frac{b}{2^n} \right)}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to \frac{\pi}{6}} \frac{\cot^2 x - 3}{cosec x - 2}\]
\[\lim_{x \to 0} \left( \cos x + \sin x \right)^{1/x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
\[\lim_{x \to 0} \frac{x}{\tan x} is\]
\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to
\[\lim_{x \to \infty} \frac{\sqrt{x^2 - 1}}{2x + 1}\]
\[\lim_{n \to \infty} \left\{ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right\}\]is equal to
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
Evaluate the following limit:
`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
Evaluate the following Limits: `lim_(x -> "a") ((x + 2)^(5/3) - ("a" + 2)^(5/3))/(x - "a")`
If the value of `lim_(x -> 1) (1 - (1 - x))^"m"/x` is 99, then n = ______.
Evaluate the Following limit:
`lim_ (x -> 3) [sqrt (x + 6)/ x]`
