Advertisements
Advertisements
प्रश्न
\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]
पर्याय
−1/12
−4/3
−16/3
−1/48
Advertisements
उत्तर
−1/48
\[\lim_{h \to 0} \left[ \frac{1}{h \sqrt[3]{8 + h}} - \frac{1}{2h} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{1}{h}\left\{ \frac{1}{\sqrt[3]{8 + h}} - \frac{1}{2} \right\} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{1}{h}\left\{ \frac{2 - \left( 8 + h \right)^{1/3}}{2 \times \sqrt[3]{8 + h}} \right\} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{1}{h}\left\{ \frac{8^{1/3} - \left( 8 + h \right)^{1/3}}{2 \sqrt[3]{8 + h}} \right\} \right] \left[ A^3 - B^3 = \left( A - B \right)\left( A^2 + AB + B^2 \right) or A - B = \frac{A^3 - B^3}{A^2 + AB + B^2} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{8 - \left( 8 + h \right)}{h\left\{ 2\sqrt[3]{8 + h} \right\}\left\{ 4 + 2 \left( 8 + h \right)^{1/3} + \left( 8 + h \right)^{2/3} \right\}} \right]\]
\[ = \left[ \frac{- 1}{2 \times \sqrt[3]{8}\left\{ 4 + 2 \times 8^{1/3} + 8^{2/3} \right\}} \right]\]
\[ = \frac{- 1}{2 \times 2\left( 4 + 4 + 4 \right)}\]
\[ = \frac{- 1}{48}\]
APPEARS IN
संबंधित प्रश्न
Suppose f(x) = `{(a+bx, x < 1),(4, x = 1),(b-ax, x > 1):}` and if `lim_(x -> 1) f(x) = f(1)` what are possible values of a and b?
\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\]
\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]
\[\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\]
\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to 0} \frac{\left( a + x \right)^2 - a^2}{x}\]
\[\lim_{x \to - 2} \frac{x^3 + x^2 + 4x + 12}{x^3 - 3x + 2}\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
\[\lim_{x \to 27} \frac{\left( x^{1/3} + 3 \right) \left( x^{1/3} - 3 \right)}{x - 27}\]
\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\]
\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\]
\[\lim_{x \to \infty} \frac{\left( 3x - 1 \right) \left( 4x - 2 \right)}{\left( x + 8 \right) \left( x - 1 \right)}\]
\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]
`lim_(x->∞) [x{sqrt(x^2+1) - sqrt(x^2-1)}]`
\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number.
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
\[\lim_{x \to 0} \frac{\sin x \cos x}{3x}\]
\[\lim_{x \to 0} \frac{5 x \cos x + 3 \sin x}{3 x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos 2x}\]
\[\lim_{x \to 0} \frac{x \cos x + \sin x}{x^2 + \tan x}\]
If \[\lim_{x \to 0} kx cosec x = \lim_{x \to 0} x cosec kx,\]
\[\lim_{x \to 0} \left( \cos x + a \sin bx \right)^{1/x}\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
\[\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + . . . + n^2}{n^3}\]
If \[f\left( x \right) = x \sin \left( 1/x \right), x \neq 0,\] then \[\lim_{x \to 0} f\left( x \right) =\]
\[\lim_{x \to 0} \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]
\[\lim_{x \to 3} \frac{\sum^n_{r = 1} x^r - \sum^n_{r = 1} 3^r}{x - 3}\]is real to
If α is a repeated root of ax2 + bx + c = 0, then \[\lim_{x \to \alpha} \frac{\tan \left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2}\]
The value of \[\lim_{x \to 0} \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} - \sqrt{a - x}}\]
\[\lim_\theta \to \pi/2 \frac{1 - \sin \theta}{\left( \pi/2 - \theta \right) \cos \theta}\] is equal to
The value of \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\]
\[\lim_{x \to \infty} \frac{\left| x \right|}{x}\] is equal to
Evaluate the Following limit:
`lim_(x->5) [(x^3 -125)/(x^5-3125)]`
Evaluate the following limit:
`\underset{x->3}{lim}[sqrt(x +6)/(x)]`
