Advertisements
Advertisements
प्रश्न
Evaluate the following limits: `lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`
Advertisements
उत्तर
`lim_(x -> 0)(root(3)(1 + x) - sqrt(1 + x))/x`
= `lim_(x -> 0)((1 + x)^(1/3) - (1 + x)^(1/2))/x`
Put 1 + x = y
As x → 0, y → 1
∴ `lim_(x -> 0)((1 + x)^(1/3) - (1 + x)^(1/2))/x`
= `lim_(y -> 1)(y^(1/3) - y^(1/2))/(y - 1)`
= `lim_(y -> 1)((y^(1/3) - 1) - (y^(1/2) - 1))/(y - 1)`
= `lim_(y -> 1)((y^(1/3) - 1)/(y - 1) - (y^(1/2) - 1)/(y - 1))`
= `lim_(y -> 1) (y^(1/3) - 1^(1/3))/(y - 1) - lim_(y -> 1)(y^(1/2) - 1^(1/2))/(y - 1)`
= `1/3(1)^((-2)/3) - 1/2(1)^((-1)/2) ...[because lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`
= `1/3 - 1/2`
= `(2 - 3)/6`
= `-1/6`
APPEARS IN
संबंधित प्रश्न
Show that \[\lim_{x \to 0} \frac{x}{\left| x \right|}\] does not exist.
\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\]
\[\lim_{x \to 2} \left( 3 - x \right)\]
\[\lim_{x \to - 1}{\left( 4 x^2 + 2 \right)}\]
\[\lim_{x \to 1} \frac{1 - x^{- 1/3}}{1 - x^{- 2/3}}\]
If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = \lim_{x \to 5} \left( 4 + x \right),\] find all possible values of a.
\[\lim_{x \to \infty} \frac{\left( 3x - 1 \right) \left( 4x - 2 \right)}{\left( x + 8 \right) \left( x - 1 \right)}\]
\[\lim_{n \to \infty} \left[ \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!} \right]\]
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos dx}\]
\[\lim_{x \to 0} \frac{\sin 3x - \sin x}{\sin x}\]
\[\lim_{x \to 0} \frac{\tan x - \sin x}{\sin 3x - 3 \sin x}\]
\[\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}\]
\[\lim_\theta \to 0 \frac{1 - \cos 4\theta}{1 - \cos 6\theta}\]
\[\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}\]
\[\lim_{x \to 1} \frac{1 - \frac{1}{x}}{\sin \pi \left( x - 1 \right)}\]
\[\lim_{x \to - 1} \frac{x^2 - x - 2}{\left( x^2 + x \right) + \sin \left( x + 1 \right)}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( 4x - \pi \right)^2}\]
The value of \[\lim_{x \to \pi/2} \left( \sec x - \tan x \right)\]is
If the value of `lim_(x -> 1) (1 - (1 - x))^"m"/x` is 99, then n = ______.
If `lim_(x -> 1) (x^4 - 1)/(x - 1) = lim_(x -> k) (x^3 - l^3)/(x^2 - k^2)`, then find the value of k.
