Advertisements
Advertisements
प्रश्न
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
Advertisements
उत्तर
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{5 + \cos \left( \pi - h \right)} - 2}{\left( \pi - \left( \pi - h \right) \right)^2}\]
\[ = \lim_{h \to 0} \frac{\sqrt{5 - \cos h} - 2}{h^2}\]
\[\text{ PRationalising the numerator, we get }: \]
\[ \lim_{h \to 0} \frac{\left( \sqrt{5 - \cos h} - 2 \right) \left( \sqrt{5 - \cos h} + 2 \right)}{h^2 \left( \sqrt{5 - \cos h} + 2 \right)}\]
\[ = \lim_{h \to 0} \frac{5 - \cos h - 4}{h^2 \left( \sqrt{5 - \cos h} + 2 \right)}\]
\[ = \lim_{h \to 0} \frac{1 - \cos h}{h^2 \left[ \sqrt{5 - \cos h} + 2 \right]}\]
\[ = \lim_{h \to 0} \frac{2 \sin^2 \frac{h}{2}}{4\left( \frac{h^2}{4} \right) \left( \sqrt{5 - \cos h} + 2 \right)}\]
\[ = \frac{1}{2 \left( \sqrt{5 - 1} + 2 \right)}\]
\[ = \frac{1}{2\left( 4 \right)}\]
\[ = \frac{1}{8}\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 1} \frac{\sqrt{x + 8}}{\sqrt{x}}\]
\[\lim_{x \to 3} \frac{x^4 - 81}{x^2 - 9}\]
\[\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\]
\[\lim_{x \to 0} \frac{\left( a + x \right)^2 - a^2}{x}\]
\[\lim_{x \to 1} \frac{x^4 - 3 x^3 + 2}{x^3 - 5 x^2 + 3x + 1}\]
\[\lim_{x \to 2} \frac{x^3 + 3 x^2 - 9x - 2}{x^3 - x - 6}\]
\[\lim_{x \to 1} \frac{1 - x^{- 1/3}}{1 - x^{- 2/3}}\]
If \[\lim_{x \to a} \frac{x^5 - a^5}{x - a} = 405,\]find all possible values of a.
If \[\lim_{x \to a} \frac{x^3 - a^3}{x - a} = \lim_{x \to 1} \frac{x^4 - 1}{x - 1},\] find all possible values of a.
\[\lim_{x \to \infty} \sqrt{x^2 + 7x - x}\]
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos dx}\]
\[\lim_\theta \to 0 \frac{\sin 3\theta}{\tan 2\theta}\]
\[\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^2 + \tan x}\]
\[\lim_{x \to 0} \frac{\sin 3x - \sin x}{\sin x}\]
\[\lim_{x \to 0} \frac{\sin 2x \left( \cos 3x - \cos x \right)}{x^3}\]
\[\lim_{x \to 0} \frac{2 \sin x^\circ - \sin 2 x^\circ}{x^3}\]
If \[\lim_{x \to 0} kx cosec x = \lim_{x \to 0} x cosec kx,\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\sin 2x}{\cos x}\]
\[\lim_{x \to a} \frac{\sin \sqrt{x} - \sin \sqrt{a}}{x - a}\]
\[\lim_{x \to 1} \frac{1 + \cos \pi x}{\left( 1 - x \right)^2}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\cos x - \sin x}{\left( \frac{\pi}{4} - x \right) \left( \cos x + \sin x \right)}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to 0} \frac{8^x - 2^x}{x}\]
\[\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n\]
\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
If \[f\left( x \right) = x \sin \left( 1/x \right), x \neq 0,\] then \[\lim_{x \to 0} f\left( x \right) =\]
\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to
\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\] is equal at
\[\lim_{x \to 1} \frac{\sin \pi x}{x - 1}\]
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
\[\lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x} - \sqrt{3}}}{x - 2}\] is equal to
The value of \[\lim_{x \to 0} \frac{1 - \cos x + 2 \sin x - \sin^3 x - x^2 + 3 x^4}{\tan^3 x - 6 \sin^2 x + x - 5 x^3}\] is
Evaluate the following limit:
`lim_(x -> 3) [sqrt(x + 6)/x]`
Evaluate the following limits: if `lim_(x -> 5)[(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.
Which of the following function is not continuous at x = 0?
Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`
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)]`
Evaluate the following limit.
`lim_(x->5)[(x^3 -125)/(x^5 - 3125)]`
