Advertisements
Advertisements
प्रश्न
\[\lim_{x \to \pi/4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}\] is equal to
पर्याय
\[\frac{1}{\sqrt{2}}\]
\[\frac{1}{2}\]
\[\frac{1}{2\sqrt{2}}\]
1
Advertisements
उत्तर
1/2
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}\cos x - 1}{\cot x - 1}\]
\[\text{ Rationalising the numerator, we get }: \]
\[ = \lim_{x \to \frac{\pi}{4}} \left( \frac{\sqrt{2} \cos x - 1}{\cot x - 1} \right) \times \left( \frac{\sqrt{2} \cos x + 1}{\sqrt{2} \cos x + 1} \right)\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\left( 2 \cos^2 x - 1 \right)}{\left( \cos x - \sin x \right)} \times \frac{\sin x}{\left( \sqrt{2}\cos x + 1 \right)}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\left( \cos^2 x - \sin^2 x \right)}{\left( \cos x - \sin x \right)} \times \frac{\sin x}{\left( \sqrt{2}\cos x + 1 \right)}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\left( \cos x + \sin x \right) \sin x}{\left( \sqrt{2}\cos x + 1 \right)}\]
\[ = \frac{\left( \cos \frac{\pi}{4} + \sin \frac{\pi}{4} \right) \sin \frac{\pi}{4}}{\left( \sqrt{2}\cos \frac{\pi}{4} + 1 \right)}\]
\[ = \frac{\left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right)\left( \frac{1}{\sqrt{2}} \right)}{\sqrt{2} . \frac{1}{\sqrt{2}} + 1}\]
\[ = \frac{\left( \frac{2}{\sqrt{2}} \right) \times \frac{1}{\sqrt{2}}}{2}\]
\[ = \frac{1}{2}\]
The correct answer is B.
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 2} \left( 3 - x \right)\]
\[\lim_{x \to - 1} \frac{x^3 - 3x + 1}{x - 1}\]
\[\lim_{x \to 3} \frac{x^2 - 4x + 3}{x^2 - 2x - 3}\]
\[\lim_{x \to 5} \frac{x^2 - 9x + 20}{x^2 - 6x + 5}\]
\[\lim_{x \to 1/4} \frac{4x - 1}{2\sqrt{x} - 1}\]
\[\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{4}{x^3 - 2 x^2} \right)\]
\[\lim_{x \to 1} \left\{ \frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3 x^2 + 2x} \right\}\]
\[\lim_{x \to - 1} \frac{x^3 + 1}{x + 1}\]
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . . n^3}{n^4} \right]\]
\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number.
Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin \left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to 0} \frac{\sin \left( a + x \right) + \sin \left( a - x \right) - 2 \sin a}{x \sin x}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x}{3 \tan^2 x}\]
\[\lim_\theta \to 0 \frac{1 - \cos 4\theta}{1 - \cos 6\theta}\]
\[\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}\]
Evaluate the following limit:
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{1 - \cos6x}}{\sqrt{2}\left( \frac{\pi}{3} - x \right)}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]
\[\lim_{x \to - 1} \frac{x^2 - x - 2}{\left( x^2 + x \right) + \sin \left( x + 1 \right)}\]
\[\lim_{x \to 1} \left( 1 - x \right) \tan \left( \frac{\pi x}{2} \right)\]
\[\lim_{x \to \frac{\pi}{2}} \left( \frac{\pi}{2} - x \right) \tan x\]
\[\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\] is equal at
\[\lim_{x \to \pi/3} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\] 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
if `lim_(x -> 2) (x^"n"- 2^"n")/(x - 2)` = 80 then find the value of n.
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 limits: `lim_(x ->3) [sqrt(x + 6)/x]`
Evaluate the following limit:
`lim_(x->5)[(x^3-125)/(x^5-3125)]`
Evaluate the following limits: `lim_(x -> 3) [sqrt(x + 6)/x]`
Evaluate the following limit:
`lim_(x->7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
