Advertisements
Advertisements
प्रश्न
Evaluate the following limits: `lim_(y -> 1) [(2y - 2)/(root(3)(7 + y) - 2)]`
उत्तर
`lim_(y -> 1) (2y - 2)/(root(3)(7 + y) - 2)`
= `lim_(y -> 1) (2(y - 1))/((7 + y)^(1/3) - 8^(1/3)) ...[because 2 = (2^3)^(1/3) = 8^(1/3)]`
= `lim_(y -> 1) 2/(((y + 7)^(1/3) - 8^(1/3))/(y - 1)`
= `(lim_(y -> 1) 2)/(lim_(y -> 1) ((y + 7)^(1/3) - 8^(1/3))/((y + 7) - 8)`
Let y + 7 = x
As y → 1, x → 8
= `2/(lim_(x -> 8) (x^(1/3) - 8^(1/3))/(x - 8))`
= `2/(1/3(8)^((-2)/2)) ...[lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`
= `2(3)*(8)^(2/3)`
= `6(2^3)^(2/3)`
= 6 x (2)2
= 24
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 0} \frac{x^{2/3} - 9}{x - 27}\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\]
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . . n^3}{n^4} \right]\]
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . n^3}{\left( n - 1 \right)^4} \right]\]
Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]
\[\lim_{x \to 0} \frac{x^2}{\sin x^2}\]
\[\lim_{x \to 0} \frac{\sin x \cos x}{3x}\]
\[\lim_\theta \to 0 \frac{\sin 3\theta}{\tan 2\theta}\]
\[\lim_{x \to 0} \frac{\sin x^2 \left( 1 - \cos x^2 \right)}{x^6}\]
\[\lim_{x \to 0} \frac{\sin 2x \left( \cos 3x - \cos x \right)}{x^3}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos 2x}\]
\[\lim_{x \to a} \frac{\cos x - \cos a}{x - a}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\left( \frac{\pi}{2} - x \right)^2}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\sqrt{2 - \sin x} - 1}{\left( \frac{\pi}{2} - x \right)^2}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{2 - {cosec}^2 x}{1 - \cot x}\]
if `lim_(x -> 2) (x^"n"- 2^"n")/(x - 2)` = 80 then find the value of n.
Evaluate the following Limit:
`lim_(x -> 0) ((1 + x)^"n" - 1)/x`
Evaluate the Following limit:
`lim_(x->3)[sqrt(x+6)/x]`
