Advertisements
Advertisements
Question
Evaluate the following limits: if `lim_(x -> 1)[(x^4 - 1)/(x - 1)] = lim_(x -> "a") [(x^3 - "a"^3)/(x - "a")]`, find all the value of a.
Advertisements
Solution
`lim_(x -> 1)[(x^4 - 1)/(x - 1)] = lim_(x -> "a") [(x^3 - "a"^3)/(x - "a")]`
`lim_(x -> 1) (x^4 - (1)^4)/(x - 1) = lim_(x -> "a") (x^3 - "a"^3)/(x - "a")`
∴ `4(1)^3 = 3"a"^2 ...[because lim_(x -> "a") (x^"n" - "a"^"n")/(x - "a") = "na"^("n" - 1)]`
∴ 3a2 = 4
∴ a2 = `4/3`
∴ a = `± 2/sqrt(3)`
APPEARS IN
RELATED QUESTIONS
\[\lim_{x \to 0} \frac{x^{2/3} - 9}{x - 27}\]
\[\lim_{x \to 2} \left( 3 - x \right)\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
If \[\lim_{x \to 3} \frac{x^n - 3^n}{x - 3} = 108,\] find the value of n.
\[\lim_{x \to \infty} \sqrt{x^2 + 7x - x}\]
\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]
Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]
\[\lim_{x \to 0} \frac{x^2 - \tan 2x}{\tan x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}\]
\[\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}\]
If \[\lim_{x \to 0} kx cosec x = \lim_{x \to 0} x cosec kx,\]
\[\lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a}\]
\[\lim_{n \to \infty} 2^{n - 1} \sin \left( \frac{a}{2^n} \right)\]
\[\lim_{x \to 1} \left( 1 - x \right) \tan \left( \frac{\pi x}{2} \right)\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to 0} \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
\[\lim_{x \to \infty} a^x \sin \left( \frac{b}{a^x} \right), a, b > 1\] is equal to
Evaluate the following limit:
`lim_(x -> 3) [sqrt(x + 6)/x]`
Evaluate the Following limit:
`lim_(x->5) [(x^3 -125)/(x^5-3125)]`
