Advertisements
Advertisements
प्रश्न
Evaluate the following limits: `lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`
उत्तर
`lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`
= `(-3). (2)^-4 ...[ lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`
= `-3 xx 1/2^4`
= `(-3)/16`
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 0} 9\]
\[\lim_{x \to 5} \frac{x^3 - 125}{x^2 - 7x + 10}\]
\[\lim_{x \to 1} \left( \frac{1}{x^2 + x - 2} - \frac{x}{x^3 - 1} \right)\]
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.
\[\lim_{x \to 0} \frac{x^2}{\sin x^2}\]
\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\]
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin \left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to 0} \frac{\tan x - \sin x}{\sin 3x - 3 \sin x}\]
\[\lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}\]
\[\lim_\theta \to 0 \frac{1 - \cos 4\theta}{1 - \cos 6\theta}\]
\[\lim_{x \to a} \frac{\cos x - \cos a}{x - a}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to \frac{3\pi}{2}} \frac{1 + {cosec}^3 x}{\cot^2 x}\]
Write the value of \[\lim_{x \to 2} \frac{\left| x - 2 \right|}{x - 2} .\]
\[\lim_{n \to \infty} \frac{n!}{\left( n + 1 \right)! + n!}\] is equal to
if `lim_(x -> 2) (x^"n"- 2^"n")/(x - 2)` = 80 then find the value of n.
Evaluate the Following limit:
`lim_(x->3)[sqrt(x+6)/x]`
Evaluate the following limit:
`lim_(x->5)[(x^3-125)/(x^5-3125)]`
Evaluate the Following limit:
`lim_ (x -> 3) [sqrt (x + 6)/ x]`
