Advertisements
Advertisements
Question
\[\lim_{x \to 0} \frac{1 - \cos 2x + \tan^2 x}{x \sin x}\]
Advertisements
Solution
\[\lim_{x \to 0} \left[ \frac{1 - \cos 2x + \tan^2 x}{x \sin x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \sin^2 x + \tan^2 x}{x \sin x} \right] \left[ \because 1 - \cos 2A = 2 \sin^2 A \right]\]
\[\text{ Dividing numerator & denominator by } x^2 : \]
\[ \lim_{x \to 0} \left[ \frac{\frac{2 \sin^2 x}{x^2} + \frac{\tan^2 x}{x^2}}{\left( \frac{\sin x}{x} \right)} \right]\]
\[ = \frac{2 \left( 1 \right)^2 + \left( 1 \right)^2}{1} \left[ \because \lim_{x \to 0} \frac{\sin^2 x}{x^2} = 1, \lim_{x \to 0} \frac{\tan^2 x}{x^2} = 1 \right]\]
\[ = 3\]
APPEARS IN
RELATED QUESTIONS
\[\lim_{x \to 1} \frac{\sqrt{x + 8}}{\sqrt{x}}\]
\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\]
\[\lim_{x \to 0} 9\]
\[\lim_{x \to 2} \left( 3 - x \right)\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{4}{x^3 - 2 x^2} \right)\]
\[\lim_{x \to 1} \frac{x^4 - 3 x^3 + 2}{x^3 - 5 x^2 + 3x + 1}\]
\[\lim_{x \to a} \frac{x^{2/3} - a^{2/3}}{x^{3/4} - a^{3/4}}\]
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.
\[\lim_{x \to \infty} \sqrt{x + 1} - \sqrt{x}\]
\[\lim_{x \to \infty} \frac{3 x^{- 1} + 4 x^{- 2}}{5 x^{- 1} + 6 x^{- 2}}\]
\[\lim_{x \to \infty} \left[ \left\{ \sqrt{x + 1} - \sqrt{x} \right\} \sqrt{x + 2} \right]\]
\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\]
\[\lim_{x \to 0} \frac{7x \cos x - 3 \sin x}{4x + \tan x}\]
\[\lim_{x \to 0} \frac{3 \sin 2x + 2x}{3x + 2 \tan 3x}\]
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin \left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to 0} \frac{2 \sin x^\circ - \sin 2 x^\circ}{x^3}\]
\[\lim_{x \to 0} \frac{5x + 4 \sin 3x}{4 \sin 2x + 7x}\]
If \[\lim_{x \to 0} kx cosec x = \lim_{x \to 0} x cosec kx,\]
\[\lim_{x \to \pi} \frac{\sin x}{\pi - x}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\cot x}{\frac{\pi}{2} - x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \sin 2x}{1 + \cos 4x}\]
\[\lim_{x \to - 1} \frac{x^2 - x - 2}{\left( x^2 + x \right) + \sin \left( x + 1 \right)}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{1 - \sqrt{2} \sin x}\]
\[\lim_{x \to \frac{3\pi}{2}} \frac{1 + {cosec}^3 x}{\cot^2 x}\]
\[\lim_{x \to 0} \left( \cos x + a \sin bx \right)^{1/x}\]
Write the value of \[\lim_{x \to 0^-} \frac{\sin \left[ x \right]}{\left[ x \right]} .\]
\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]
The value of \[\lim_{x \to \infty} \frac{\sqrt{1 + x^4} + \left( 1 + x^2 \right)}{x^2}\] is
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\] is equal to
If \[f\left( x \right) = \left\{ \begin{array}{l}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0\end{array}, \right.\] then \[\lim_{x \to 0} f\left( x \right)\] equals
\[\lim_{n \to \infty} \frac{n!}{\left( n + 1 \right)! + n!}\] is equal to
\[\lim_{x \to 0} \frac{8}{x^8}\left\{ 1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cos \frac{x^2}{4} \right\}\] 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
The value of \[\lim_{x \to \infty} \frac{\left( x + 1 \right)^{10} + \left( x + 2 \right)^{10} + . . . + \left( x + 100 \right)^{10}}{x^{10} + {10}^{10}}\] is
\[\lim_{x \to 1} \left[ x - 1 \right]\] where [.] is the greatest integer function, is equal to
Evaluate the following limit:
`lim_(x -> 7)[((root(3)(x) - root(3)(7))(root(3)(x) + root(3)(7)))/(x - 7)]`
`lim_(x->3) (x^5 - 243)/(x^3 - 27)` = ?
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->7)[((root(3)(x)-root(3)(7))(root(3)(x)+root(3)(7)))/(x-7)]`
