Advertisements
Advertisements
Question
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( \frac{\pi}{4} - x \right)^2}\]
Advertisements
Solution
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( \frac{\pi}{4} - x \right)^2}\]
\[\text{ Dividing the numerator and the denominator by }\sqrt{2}:\]
\[ \lim_{x \to \frac{\pi}{4}} \frac{1 - \left( \frac{1}{\sqrt{2}}\cos x + \frac{1}{\sqrt{2}}\sin x \right)}{\frac{1}{\sqrt{2}} \left( \frac{\pi}{4} - x \right)^2}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{1 - \left( \cos \frac{\pi}{4}\cos x + \sin \frac{\pi}{4} \sin x \right)}{\frac{1}{\sqrt{2}} \left( \frac{\pi}{4} - x \right)^2}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2}\left( 1 - \cos \left( \frac{\pi}{4} - x \right) \right)}{\left( \frac{\pi}{4} - x \right)^2}\]
\[ = \lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} \left( 2 \sin^2 \left( \frac{\frac{\pi}{4} - x}{2} \right) \right)}{\left( \frac{\pi}{4} - x \right)^2} \left[ \because 1 - \cos \theta = 2 \sin \frac{2\theta}{2} \right]\]
\[ = 2\sqrt{2} \lim_{x \to \frac{\pi}{4}} \frac{\left[ \sin^2 \left( \frac{\frac{\pi}{4} - x}{2} \right) \right]}{4 \left( \frac{\frac{\pi}{4} - x}{2} \right)^2}\]
\[ = \frac{2\sqrt{2}}{4}\]
\[ = \frac{1}{\sqrt{2}}\]
APPEARS IN
RELATED QUESTIONS
\[\lim_{x \to 0} \frac{2 x^2 + 3x + 4}{x^2 + 3x + 2}\]
\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\]
\[\lim_{x \to - 5} \frac{2 x^2 + 9x - 5}{x + 5}\]
\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to 0} \frac{\left( a + x \right)^2 - a^2}{x}\]
\[\lim_{x \to 1} \left( \frac{1}{x - 1} - \frac{2}{x^2 - 1} \right)\]
\[\lim_{x \to 3} \left( x^2 - 9 \right) \left[ \frac{1}{x + 3} + \frac{1}{x - 3} \right]\]
\[\lim_{x \to 3} \frac{x^2 - x - 6}{x^3 - 3 x^2 + x - 3}\]
\[\lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x^3 - 3 x^2 + 2x} \right]\]
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
\[\lim_{x \to a} \frac{x^{5/7} - a^{5/7}}{x^{2/7} - a^{2/7}}\]
\[\lim_{x \to \infty} \frac{5 x^3 - 6}{\sqrt{9 + 4 x^6}}\]
\[\lim_{x \to - \infty} \left( \sqrt{x^2 - 8x} + x \right)\]
\[\lim_{x \to 0} \frac{x^2}{\sin x^2}\]
\[\lim_{x \to 0} \frac{\sin x^2 \left( 1 - \cos x^2 \right)}{x^6}\]
\[\lim_{x \to 0} \frac{\sin^2 4 x^2}{x^4}\]
\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{x^3 \cot x}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos 2x}\]
\[\lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}\]
\[\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}\]
Evaluate the following limit:
\[\lim_{x \to 0} \frac{\sin\left( \alpha + \beta \right)x + \sin\left( \alpha - \beta \right)x + \sin2\alpha x}{\cos^2 \beta x - \cos^2 \alpha x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
\[\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n\]
\[\lim_{x \to 0} \left( \cos x \right)^{1/\sin x}\]
\[\lim_{x \to 0} \left( \cos x + \sin x \right)^{1/x}\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
\[\lim_{n \to \infty} \left\{ \frac{1}{1 - n^2} + \frac{2}{1 - n^2} + . . . + \frac{n}{1 - n^2} \right\}\]
\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to
\[\lim_{x \to 3} \frac{\sum^n_{r = 1} x^r - \sum^n_{r = 1} 3^r}{x - 3}\]is real to
\[\lim_{n \to \infty} \frac{n!}{\left( n + 1 \right)! + n!}\] is equal to
If \[f\left( x \right) = \begin{cases}\frac{\sin\left[ x \right]}{\left[ x \right]}, & \left[ x \right] \neq 0 \\ 0, & \left[ x \right] = 0\end{cases}\] where denotes the greatest integer function, then \[\lim_{x \to 0} f\left( x \right)\]
Evaluate the following limit:
`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
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]`
Evaluate the Following limit:
`lim_ (x -> 3) [sqrt (x + 6)/ x]`
