Advertisements
Advertisements
Question
Evaluate the following limits: `lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`
Advertisements
Solution
`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
RELATED QUESTIONS
Show that \[\lim_{x \to 0} \frac{x}{\left| x \right|}\] does not exist.
\[\lim_{x \to - 5} \frac{2 x^2 + 9x - 5}{x + 5}\]
\[\lim_{x \to 3} \frac{x^2 - x - 6}{x^3 - 3 x^2 + x - 3}\]
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
\[\lim_{x \to \infty} \frac{\left( 3x - 1 \right) \left( 4x - 2 \right)}{\left( x + 8 \right) \left( x - 1 \right)}\]
\[\lim_{x \to \infty} \frac{5 x^3 - 6}{\sqrt{9 + 4 x^6}}\]
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos dx}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}\]
\[\lim_{x \to 0} \frac{\tan 2x - \sin 2x}{x^3}\]
\[\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a}\]
\[\lim_{x \to - 1} \frac{x^2 - x - 2}{\left( x^2 + x \right) + \sin \left( x + 1 \right)}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 0^-} \frac{\sin x}{\sqrt{x}} .\]
The value of \[\lim_{x \to 0} \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} - \sqrt{a - x}}\]
The value of \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\]
`lim_(x->3) (x^5 - 243)/(x^3 - 27)` = ?
If `f(x) = {{:(x + 2",", x ≤ - 1),(cx^2",", x > -1):}`, find 'c' if `lim_(x -> -1) f(x)` exists
Evaluate the Following limit:
`lim_(x->5) [(x^3 -125)/(x^5-3125)]`
