Advertisements
Advertisements
Question
Let `f(x) = {{:((k cos x)/(pi - 2x)",", "when" x ≠ pi/2),(3",", x = pi/2 "and if" f(x) = f(pi/2)):}` find the value of k.
Advertisements
Solution
Given, `f(x) = {{:((k cos x)/(pi - 2x)",", x ≠ pi/2),(3",", x = pi/2):}`
L.H.L, `f(x) = lim_(x -> pi^-/2) (k cos x)/(pi - 2x)`
= `lim_(h -> 0) (k cos (pi/2 - h))/(pi - 2(pi/2 - h))`
= `lim_(h -> 0) (k sin h)/(pi - pi + 2h)`
= `lim_(h -> 0) (k sin h)/(2h)`
= `l/2 * 1`
= `k/2` ......`[because lim_(h -> 0) sinh/h = 1]`
R.H.L. `f(x) = lim_(x -> pi^+/2) (k cos x)/(pi - 2x)`
= `lim_(h -> 0) (k cos (pi/2 + h))/(pi - 2(pi/2 + h))`
= `lim_(h -> 0) (-k sin h)/(pi - pi - 2h)`
= `lim_(h -> 0) (-k sin h)/(-2h)`
= `k/2` ....`[because lim_(h -> 0) sinh/h = 1]`
We are given that `lim_(x -> pi/2) f(x)` = 3
So, `k/2` = 3
⇒ k = 6
APPEARS IN
RELATED QUESTIONS
\[\lim_{x \to - 1} \frac{x^3 - 3x + 1}{x - 1}\]
\[\lim_{x \to 4} \frac{x^2 - 7x + 12}{x^2 - 3x - 4}\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to 1/4} \frac{4x - 1}{2\sqrt{x} - 1}\]
\[\lim_{x \to 0} \frac{\left( a + x \right)^2 - a^2}{x}\]
\[\lim_{x \to 1} \frac{1 - x^{- 1/3}}{1 - x^{- 2/3}}\]
\[\lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x^3 - 3 x^2 + 2x} \right]\]
\[\lim_{x \to 1} \frac{\sqrt{x^2 - 1} + \sqrt{x - 1}}{\sqrt{x^2 - 1}}, x > 1\]
\[\lim_{x \to \infty} \frac{x}{\sqrt{4 x^2 + 1} - 1}\]
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . . n^3}{n^4} \right]\]
\[f\left( x \right) = \frac{a x^2 + b}{x^2 + 1}, \lim_{x \to 0} f\left( x \right) = 1\] and \[\lim_{x \to \infty} f\left( x \right) = 1,\]then prove that f(−2) = f(2) = 1
Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\]
\[\lim_{x \to 0} \frac{x^2}{\sin x^2}\]
\[\lim_{x \to 0} \frac{\cos 3x - \cos 7x}{x^2}\]
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin \left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to 0} \frac{\sec 5x - \sec 3x}{\sec 3x - \sec x}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x + \tan^2 x}{x \sin x}\]
\[\lim_\theta \to 0 \frac{1 - \cos 4\theta}{1 - \cos 6\theta}\]
\[\lim_\theta \to 0 \frac{\sin 4\theta}{\tan 3\theta}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{1 - \sqrt{2} \sin x}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log a}{x}\]
\[\lim_{x \to 0^-} \frac{\sin \left[ x \right]}{\left[ x \right]} .\]
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]
Write the value of \[\lim_{x \to 0^-} \left[ x \right] .\]
Write the value of \[\lim_{x \to 2} \frac{\left| x - 2 \right|}{x - 2} .\]
\[\lim_{x \to 0^-} \frac{\sin x}{\sqrt{x}} .\]
\[\lim_{x \to \infty} \frac{\sqrt{x^2 - 1}}{2x + 1}\]
\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\] 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
\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] is equal to
\[\lim_\theta \to \pi/2 \frac{1 - \sin \theta}{\left( \pi/2 - \theta \right) \cos \theta}\] 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 limits: if `lim_(x -> 5)[(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.
Evaluate `lim_(h -> 0) ((a + h)^2 sin (a + h) - a^2 sina)/h`
