Advertisements
Advertisements
Question
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{1 - \sqrt{2} \sin x}\]
Advertisements
Solution
\[\lim_{x \to \frac{\pi}{4}} \left[ \frac{1 - \tan x}{1 - \sqrt{2} \sin x} \right]\]
\[\text{ It is of } \frac{0}{0} \text{ form } .\]
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
RELATED QUESTIONS
\[\lim_{x \to 0} \frac{3x + 1}{x + 3}\]
\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\]
\[\lim_{x \to 5} \frac{x^2 - 9x + 20}{x^2 - 6x + 5}\]
\[\lim_{x \to \sqrt{2}} \frac{x^2 - 2}{x^2 + \sqrt{2}x - 4}\]
\[\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
If \[\lim_{x \to 3} \frac{x^n - 3^n}{x - 3} = 108,\] find the value of n.
\[\lim_{x \to \infty} \sqrt{x^2 + cx - x}\]
Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]
\[\lim_{x \to 0} \frac{\sin x^0}{x}\]
\[\lim_{x \to 0} \left[ \frac{x^2}{\sin x^2} \right]\]
\[\lim_{x \to 0} \frac{\sin x^n}{x^n}\]
\[\lim_{x \to 0} \frac{\tan 3x - 2x}{3x - \sin^2 x}\]
\[\lim_{x \to 0} \frac{\tan x - \sin x}{\sin 3x - 3 \sin x}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x}{3 \tan^2 x}\]
\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]
\[\lim_{x \to 0} \left( cosec x - \cot x \right)\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log a}{x}\]
\[\lim_{x \to 0} \left( \cos x + \sin x \right)^{1/x}\]
\[\lim_{x \to 0^-} \frac{\sin \left[ x \right]}{\left[ x \right]} .\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
\[\lim_{x \to \infty} \frac{\sin x}{x}\] equals
\[\lim_{n \to \infty} \left\{ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right\}\]is equal to
The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\] is
\[\lim_{n \to \infty} \frac{n!}{\left( n + 1 \right)! + n!}\] is equal to
\[\lim_{x \to \infty} a^x \sin \left( \frac{b}{a^x} \right), a, b > 1\] is equal to
If α is a repeated root of ax2 + bx + c = 0, then \[\lim_{x \to \alpha} \frac{\tan \left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2}\]
Evaluate the following limits: `lim_(x -> 0)[(root(3)(1 + x) - sqrt(1 + x))/x]`
Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^2 - 25)]`
Evaluate the following Limit:
`lim_(x -> 0) ((1 + x)^"n" - 1)/x`
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.
Evaluate the following limit:
`lim_(x->7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
