Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\]
Advertisements
उत्तर
\[\lim_{x \to 0} \left[ \frac{\tan 8x}{\sin 2x} \right]\]
\[\Rightarrow \lim_{x \to 0} \left[ \frac{\tan 8x}{8x} \times \frac{8x}{\frac{\sin 2x}{2x} \times 2x} \right] \left[ \because \lim_{x \to 0} \frac{\tan 8x}{8x} = 1, \lim_{x \to 0} \frac{\sin 2x}{2x} = 1 \right]\]
\[ \Rightarrow 4\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 0} \frac{3x + 1}{x + 3}\]
\[\lim_{x \to 2} \frac{x^4 - 16}{x - 2}\]
\[\lim_{x \to \sqrt{3}} \frac{x^2 - 3}{x^2 + 3 \sqrt{3}x - 12}\]
\[\lim_{x \to 1} \left( \frac{1}{x^2 + x - 2} - \frac{x}{x^3 - 1} \right)\]
\[\lim_{x \to 1} \left\{ \frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3 x^2 + 2x} \right\}\]
\[\lim_{x \to \infty} \frac{3 x^3 - 4 x^2 + 6x - 1}{2 x^3 + x^2 - 5x + 7}\]
\[\lim_{x \to \infty} \sqrt{x^2 + cx - x}\]
\[\lim_{x \to \infty} \left[ \left\{ \sqrt{x + 1} - \sqrt{x} \right\} \sqrt{x + 2} \right]\]
\[\lim_{n \to \infty} \left[ \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + . . . + \frac{1}{3^n} \right]\]
\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\]
\[\lim_{x \to 0} \frac{\sin x^n}{x^n}\]
\[\lim_{x \to 0} \frac{7x \cos x - 3 \sin x}{4x + \tan x}\]
\[\lim_{x \to 0} \frac{\sin x^2 \left( 1 - \cos x^2 \right)}{x^6}\]
\[\lim_{x \to 0} \frac{\sin \left( a + x \right) + \sin \left( a - x \right) - 2 \sin a}{x \sin x}\]
\[\lim_{x \to 0} \frac{x^2 + 1 - \cos x}{x \sin x}\]
\[\lim_{x \to 0} \frac{2 \sin x^\circ - \sin 2 x^\circ}{x^3}\]
\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]
Evaluate the following limits:
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - 1}\]
\[\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\left( \frac{\pi}{2} - x \right)^2}\]
\[\lim_{x \to \pi} \frac{\sqrt{5 + \cos x} - 2}{\left( \pi - x \right)^2}\]
Evaluate the following limit:
\[\lim_{x \to \pi} \frac{1 - \sin\frac{x}{2}}{\cos\frac{x}{2}\left( \cos\frac{x}{4} - \sin\frac{x}{4} \right)}\]
\[\lim_{x \to 0} \left( \cos x + a \sin bx \right)^{1/x}\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
Write the value of \[\lim_{x \to 2} \frac{\left| x - 2 \right|}{x - 2} .\]
Write the value of \[\lim_{x \to 0} \frac{\sin x^\circ}{x} .\]
\[\lim_{n \to \infty} \frac{1^2 + 2^2 + 3^2 + . . . + n^2}{n^3}\]
\[\lim_{x \to \infty} \frac{\sin x}{x}\] equals
\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\] is equal at
\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\] is equal to
\[\lim_{x \to \pi/3} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\] 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
\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] 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 \[\lim_{n \to \infty} \frac{\left( n + 2 \right)! + \left( n + 1 \right)!}{\left( n + 2 \right)! - \left( n + 1 \right)!}\] is
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: `lim_(x -> 5)[(x^3 - 125)/(x^2 - 25)]`
Evaluate the following limit:
`lim _ (x -> 5) [(x^3 - 125) / (x^5 - 3125)]`
Evaluate the following limit:
`\underset{x->3}{lim}[sqrt(x +6)/(x)]`
