Advertisements
Advertisements
Question
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
Advertisements
Solution
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{5 + \cos \left( \pi - h \right)} - 2}{\left( \pi - \left( \pi - h \right) \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{5 - \cos h} - 2}{h^2}\]
\[\text{ PRationalising the numerator, we get }: \]
\[ \lim_{h \to 0} \frac{\left( \sqrt{5 - \cos h} - 2 \right) \left( \sqrt{5 - \cos h} + 2 \right)}{h^2 \left( \sqrt{5 - \cos h} + 2 \right)}\]
\[ = \lim_{h \to 0} \frac{5 - \cos h - 4}{h^2 \left( \sqrt{5 - \cos h} + 2 \right)}\]
\[ = \lim_{h \to 0} \frac{1 - \cos h}{h^2 \left[ \sqrt{5 - \cos h} + 2 \right]}\]
\[ = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2}}{4\left( \frac{h^2}{4} \right) \left( \sqrt{5 - \cos h} + 2 \right)}\]
\[ = \frac{1}{2 \left( \sqrt{5 - 1} + 2 \right)}\]
\[ = \frac{1}{2\left( 4 \right)}\]
\[ = \frac{1}{8}\]
APPEARS IN
RELATED QUESTIONS
\[\lim_{x \to 1/4} \frac{4x - 1}{2\sqrt{x} - 1}\]
\[\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\]
\[\lim_{x \to 1} \frac{1 - x^{- 1/3}}{1 - x^{- 2/3}}\]
\[\lim_{x \to 0} \frac{\left( 1 + x \right)^6 - 1}{\left( 1 + x \right)^2 - 1}\]
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
If \[\lim_{x \to a} \frac{x^9 - a^9}{x - a} = 9,\] find all possible values of a.
\[\lim_{x \to \infty} \frac{\left( 3x - 1 \right) \left( 4x - 2 \right)}{\left( x + 8 \right) \left( x - 1 \right)}\]
\[\lim_{x \to \infty} \frac{x}{\sqrt{4 x^2 + 1} - 1}\]
\[\lim_{n \to \infty} \frac{n^2}{1 + 2 + 3 + . . . + n}\]
\[\lim_{n \to \infty} \left[ \frac{1^2 + 2^2 + . . . + n^2}{n^3} \right]\]
\[\lim_{n \to \infty} \left[ \frac{1 + 2 + 3 . . . . . . n - 1}{n^2} \right]\]
\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . n^3}{\left( n - 1 \right)^4} \right]\]
\[\lim_{x \to 0} \left[ \frac{x^2}{\sin x^2} \right]\]
\[\lim_{x \to 0} \frac{3 \sin 2x + 2x}{3x + 2 \tan 3x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x}\]
Evaluate the following limits:
\[\lim_{x \to 0} \frac{2\sin x - \sin2x}{x^3}\]
\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]
If \[\lim_{x \to 0} kx cosec x = \lim_{x \to 0} x cosec kx,\]
Evaluate the following limit:
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{1 - \cos6x}}{\sqrt{2}\left( \frac{\pi}{3} - x \right)}\]
\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
\[\lim_{x \to \frac{\pi}{8}} \frac{\cot 4x - \cos 4x}{\left( \pi - 8x \right)^3}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin 2\pi x}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
Write the value of \[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]
\[\lim_{x \to 0^-} \frac{\sin \left[ x \right]}{\left[ x \right]} .\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
Write the value of \[\lim_{x \to 0^-} \frac{\sin \left[ x \right]}{\left[ x \right]} .\]
\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]
\[\lim_{n \to \infty} \left\{ \frac{1}{1 . 3} + \frac{1}{3 . 5} + \frac{1}{5 . 7} + . . . + \frac{1}{\left( 2n + 1 \right) \left( 2n + 3 \right)} \right\}\]is equal to
The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\] is
\[\lim_{x \to 3} \frac{\sum^n_{r = 1} x^r - \sum^n_{r = 1} 3^r}{x - 3}\]is real to
Evaluate the following limits: if `lim_(x -> 1)[(x^4 - 1)/(x - 1)] = lim_(x -> "a") [(x^3 - "a"^3)/(x - "a")]`, find all the value of a.
Evaluate the following limits: `lim_(x -> 0)[((1 - x)^8 - 1)/((1 - x)^2 - 1)]`
Evaluate the following limits: `lim_(x -> "a")[((z + 2)^(3/2) - ("a" + 2)^(3/2))/(z - "a")]`
Which of the following function is not continuous at x = 0?
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.
Evaluate the Following limit:
`lim_(x->5) [(x^3 -125)/(x^5-3125)]`
Evaluate the following limits: `lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`
