Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}\]
उत्तर
\[\lim_{x \to 0} \left[ \frac{1 - \cos \left( 4x \right)}{x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \sin^2 2x}{x^2} \right]\]
\[ = \lim_{x \to 0} 2\left[ \frac{\sin 2x}{x} \times \frac{\sin 2x}{x} \right]\]
\[ = \lim_{x \to 0} \left[ 2 \times \frac{\sin 2x}{2x} \times \frac{\sin2x}{2x} \times 4 \right]\]
\[ = 2 \times 1 \times 1 \times 4\]
\[ = 8\]
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?
\[\lim_{x \to 1} \frac{1 + \left( x - 1 \right)^2}{1 + x^2}\]
\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]
\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\]
\[\lim_{x \to 1} \frac{x^3 + 3 x^2 - 6x + 2}{x^3 + 3 x^2 - 3x - 1}\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\]
\[\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}\]
\[\lim_{x \to a} \frac{x^{2/3} - a^{2/3}}{x^{3/4} - a^{3/4}}\]
If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = 9,\] find all possible values of a.
If \[\lim_{x \to a} \frac{x^3 - a^3}{x - a} = \lim_{x \to 1} \frac{x^4 - 1}{x - 1},\] find all possible values of a.
\[\lim_{x \to \infty} \left[ \left\{ \sqrt{x + 1} - \sqrt{x} \right\} \sqrt{x + 2} \right]\]
\[\lim_{x \to - \infty} \left( \sqrt{x^2 - 8x} + x \right)\]
Evaluate: \[\lim_{n \to \infty} \frac{1^4 + 2^4 + 3^4 + . . . + n^4}{n^5} - \lim_{n \to \infty} \frac{1^3 + 2^3 + . . . + n^3}{n^5}\]
\[\lim_{x \to 0} \frac{\sin x \cos x}{3x}\]
\[\lim_{x \to 0} \frac{\sin 3x - \sin x}{\sin x}\]
\[\lim_{x \to 0} \frac{\sin 5x - \sin 3x}{\sin x}\]
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin \left( a + h \right) - a^2 \sin a}{h}\]
Evaluate the following limit:
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin\left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{1 - \sin x}\]
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\cot x}{\frac{\pi}{2} - x}\]
\[\lim_{x \to \frac{\pi}{8}} \frac{\cot 4x - \cos 4x}{\left( \pi - 8x \right)^3}\]
\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
\[\lim_{x \to \frac{\pi}{6}} \frac{\cot^2 x - 3}{cosec x - 2}\]
Write the value of \[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] is equal to
\[\lim_{x \to 0} \frac{8}{x^8}\left\{ 1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cos \frac{x^2}{4} \right\}\] is equal to
The value of \[\lim_{x \to 0} \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a + x} - \sqrt{a - x}}\]
Evaluate the following limit:
`lim_(x -> 3) [sqrt(x + 6)/x]`
Evaluate the following limits: if `lim_(x -> 5)[(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.
Evaluate the following limits: `lim_(y -> 1) [(2y - 2)/(root(3)(7 + y) - 2)]`
if `lim_(x -> 2) (x^"n"- 2^"n")/(x - 2)` = 80 then find the value of n.
Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`
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.
If f(x) = `{{:(1 if x "is rational"),(-1 if x "is rational"):}` is continuous on ______.
Evaluate the Following limit:
`lim_ (x -> 3) [sqrt (x + 6)/ x]`
