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
Show that \[\lim_{x \to 0} \frac{x}{\left| x \right|}\] does not exist.
\[\lim_{x \to 1} \frac{x^2 + 1}{x + 1}\]
\[\lim_{x \to 0} 9\]
\[\lim_{x \to - 1}{\left( 4 x^2 + 2 \right)}\]
\[\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\]
\[\lim_{x \to a} \frac{x^{2/3} - a^{2/3}}{x^{3/4} - a^{3/4}}\]
\[\lim_{x \to 0} \frac{\sin x^0}{x}\]
\[\lim_{x \to 0} \left[ \frac{x^2}{\sin x^2} \right]\]
\[\lim_{x \to 0} \frac{\tan mx}{\tan nx}\]
\[\lim_{x \to 0} \frac{7x \cos x - 3 \sin x}{4x + \tan x}\]
\[\lim_{x \to 0} \frac{\tan^2 3x}{x^2}\]
\[\lim_{x \to 0} \frac{1 - \cos mx}{x^2}\]
\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\]
\[\lim_{x \to 0} \frac{\tan x - \sin x}{\sin 3x - 3 \sin x}\]
\[\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^2}\]
\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]
Evaluate the following limit:
\[\lim_{x \to 0} \frac{\sin\left( \alpha + \beta \right)x + \sin\left( \alpha - \beta \right)x + \sin2\alpha x}{\cos^2 \beta x - \cos^2 \alpha x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\cot x}{\frac{\pi}{2} - x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( \frac{\pi}{4} - x \right)^2}\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a}\]
\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( 4x - \pi \right)^2}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
\[\lim_{x \to } \frac{1 - \cos 2x}{x} is\]
\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]
\[\lim_{n \to \infty} \frac{n!}{\left( n + 1 \right)! + n!}\] is equal to
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_(x -> "a") ((x + 2)^(5/3) - ("a" + 2)^(5/3))/(x - "a")`
`1/(ax^2 + bx + c)`
Evaluate the following limits: `lim_(x -> 3) [sqrt(x + 6)/x]`
Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
Evaluate the following limit.
`lim_(x->3)[sqrt(x + 6)/x]`
