Advertisements
Advertisements
प्रश्न
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{1 - \sqrt{2} \sin x}\]
Advertisements
उत्तर
\[\lim_{x \to \frac{\pi}{4}} \left[ \frac{1 - \tan x}{1 - \sqrt{2} \sin x} \right]\]
\[\text{ It is of } \frac{0}{0} \text{ form } .\]
Rationalising the denominator, we get:
\[\lim_{x \to \frac{\pi}{4}} \left[ \frac{\left( 1 - \tan x \right) \left( 1 + \sqrt{2} \sin x \right)}{\left( 1 - \sqrt{2} \sin x \right) \left( 1 + \sqrt{2} \sin x \right)} \right]\]
\[ = \lim_{x \to \frac{\pi}{4}} \left[ \frac{\left( 1 - \tan x \right) \left( 1 + \sqrt{2} \sin x \right)}{1 - 2 \sin^2 x} \right]\]
\[ = \lim_{x \to \frac{\pi}{4}} \left[ \frac{\left( 1 - \frac{\sin x}{\cos x} \right) \left( 1 + \sqrt{2} \sin x \right)}{\cos 2x} \right]\]
\[ = \lim_{x \to \frac{\pi}{4}} \left[ \frac{\left( \cos x - \sin x \right) \left( 1 + \sqrt{2} \sin x \right)}{\cos x \cos 2x} \right] \]
\[ = \lim_{x \to \frac{\pi}{4}} \left[ \frac{\left( \cos x - \sin x \right) \left( 1 + \sqrt{2} \sin x \right)}{\cos x \cdot \left( \cos^2 x - \sin^2 x \right)} \right]\]
\[ = \lim_{x \to \frac{\pi}{4}} \left[ \frac{\left( \cos x - \sin x \right) \left( 1 + \sqrt{2} \sin x \right)}{\cos x \left[ \cos x - \sin x \right] \left[ \cos x + \sin x \right]} \right]\]
\[ = \frac{\left( 1 + \sqrt{2} \times \frac{1}{\sqrt{2}} \right)}{\left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right)}\]
\[ = \frac{2}{\frac{1}{\sqrt{2}} \times \sqrt{2}}\]
\[ = 2\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\]
\[\lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}\]
\[\lim_{x \to a} \frac{x^{2/3} - a^{2/3}}{x^{3/4} - a^{3/4}}\]
\[\lim_{x \to \infty} \sqrt{x + 1} - \sqrt{x}\]
\[\lim_{x \to \infty} \sqrt{x^2 + 7x - x}\]
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . . n^3}{n^4} \right]\]
Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
\[\lim_{x \to - \infty} \left( \sqrt{x^2 - 8x} + x \right)\]
\[\lim_{x \to 0} \frac{3 \sin x - 4 \sin^3 x}{x}\]
\[\lim_{x \to 0} \frac{\sin x^n}{x^n}\]
\[\lim_{x \to 0} \frac{3 \sin 2x + 2x}{3x + 2 \tan 3x}\]
\[\lim_{x \to 0} \frac{\tan 3x - 2x}{3x - \sin^2 x}\]
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin \left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]
\[\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^2}\]
\[\lim_{x \to 0} \frac{3 \sin^2 x - 2 \sin x^2}{3 x^2}\]
\[\lim_{x \to 0} \frac{1 - \cos 4x}{x^2}\]
\[\lim_{x \to 0} \frac{x \cos x + \sin x}{x^2 + \tan x}\]
Evaluate the following limits:
\[\lim_{x \to 0} \frac{2\sin x - \sin2x}{x^3}\]
\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\sqrt{2 - \sin x} - 1}{\left( \frac{\pi}{2} - x \right)^2}\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{f\left( x \right) - f\left( \frac{\pi}{4} \right)}{x - \frac{\pi}{4}},\]
\[\lim_{x \to 1} \left( 1 - x \right) \tan \left( \frac{\pi x}{2} \right)\]
\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\]
\[\lim_{x \to 0^+} \left\{ 1 + \tan^2 \sqrt{x} \right\}^{1/2x}\]
\[\lim_{x \to 0} \left( \cos x + a \sin bx \right)^{1/x}\]
Write the value of \[\lim_{x \to \infty} \frac{\sin x}{x} .\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
\[\lim_{x \to \pi/4} \frac{\sqrt{2} \cos x - 1}{\cot x - 1}\] is equal to
\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]
\[\lim_{x \to \pi/3} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\] is equal to
Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
if `lim_(x -> 2) (x^"n"- 2^"n")/(x - 2)` = 80 then find the value of n.
Evaluate the following Limits: `lim_(x -> "a") ((x + 2)^(5/3) - ("a" + 2)^(5/3))/(x - "a")`
Evaluate `lim_(h -> 0) ((a + h)^2 sin (a + h) - a^2 sina)/h`
`1/(ax^2 + bx + c)`
