Advertisements
Advertisements
प्रश्न
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.
उत्तर
Given that `lim_(x -> 1) (x^4 - 1)/(x - 1) = lim_(x -> k) (x^3 - l^3)/(x^2 - k^2)`
⇒ `4(1)^(4 - 1) = lim_(x -> k) ((x - k)(x^2 + k^2 + kx))/((x - k)(x + k))`
⇒ 4 = `lim_(x -> k) (x^2 + k^2 + kx)/(x + k)`
⇒ 4 = `(k^2 + k^2 + k^2)/(2k)`
⇒ 4 = `(3k^2)/(2k)`
⇒ 4 = `3/2 k`
⇒ k = `8/3`
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 1} \frac{\sqrt{x + 8}}{\sqrt{x}}\]
\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
\[\lim_{x \to 1} \left( \frac{1}{x - 1} - \frac{2}{x^2 - 1} \right)\]
Evaluate the following limit:
\[\lim_{x \to 1} \frac{x^7 - 2 x^5 + 1}{x^3 - 3 x^2 + 2}\]
\[\lim_{x \to a} \frac{x^{2/7} - a^{2/7}}{x - a}\]
\[\lim_{x \to 27} \frac{\left( x^{1/3} + 3 \right) \left( x^{1/3} - 3 \right)}{x - 27}\]
\[\lim_{x \to a} \frac{x^{2/3} - a^{2/3}}{x^{3/4} - a^{3/4}}\]
If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = 9,\] find all possible values of a.
\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]
\[\lim_{x \to - \infty} \left( \sqrt{x^2 - 8x} + x \right)\]
\[\lim_{x \to 0} \frac{3 \sin 2x + 2x}{3x + 2 \tan 3x}\]
\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\]
\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos 2x}\]
\[\lim_{n \to \infty} 2^{n - 1} \sin \left( \frac{a}{2^n} \right)\]
\[\lim_{x \to \frac{\pi}{4}} \frac{{cosec}^2 x - 2}{\cot x - 1}\]
\[\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n\]
\[\lim_{x \to 0} \left( \cos x + \sin x \right)^{1/x}\]
\[\lim_{x \to 0} \left( \cos x + a \sin bx \right)^{1/x}\]
\[\lim_{x \to \infty} \frac{\sin x}{x} .\]
\[\lim_{x \to \infty} \frac{\sin x}{x}\] equals
\[\lim_{n \to \infty} \left\{ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right\}\]is equal to
The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\] is
\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to
The value of \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\]
Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^2 - 25)]`