Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 0} \frac{\sin 5x}{\tan 3x}\]
उत्तर
\[\lim_{x \to 0} \left[ \frac{\sin 5x}{\tan 3x} \right]\]
\[\Rightarrow \lim_{x \to 0} \left[ \lim_{x \to 0} \frac{\sin 5x}{5x} \times \frac{5x}{\frac{\tan 3x}{3x} \times 3x} \right]\]
\[ \Rightarrow \frac{5}{3} \left[ \because \lim_{x \to 0} \frac{\sin x}{x} = 1, \lim_{x \to 0} \frac{\tan x}{x} = 1 \right]\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\]
\[\lim_{x \to - 1} \frac{x^3 - 3x + 1}{x - 1}\]
\[\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\]
\[\lim_{x \to - 2} \frac{x^3 + x^2 + 4x + 12}{x^3 - 3x + 2}\]
\[\lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x^3 - 3 x^2 + 2x} \right]\]
\[\lim_{x \to \infty} \frac{x}{\sqrt{4 x^2 + 1} - 1}\]
\[\lim_{n \to \infty} \left[ \frac{1 + 2 + 3 . . . . . . n - 1}{n^2} \right]\]
\[\lim_{n \to \infty} \left[ \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + . . . + \frac{1}{3^n} \right]\]
\[f\left( x \right) = \frac{a x^2 + b}{x^2 + 1}, \lim_{x \to 0} f\left( x \right) = 1\] and \[\lim_{x \to \infty} f\left( x \right) = 1,\]then prove that f(−2) = f(2) = 1
Evaluate: \[\lim_{n \to \infty} \frac{1^4 + 2^4 + 3^4 + . . . + n^4}{n^5} - \lim_{n \to \infty} \frac{1^3 + 2^3 + . . . + n^3}{n^5}\]
\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\]
\[\lim_{x \to 0} \frac{\tan^2 3x}{x^2}\]
\[\lim_{x \to 0} \frac{1 - \cos 2x}{\cos 2x - \cos 8x}\]
\[\lim_{x \to 0} \frac{x^3 \cot x}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{\sin \left( 3 + x \right) - \sin \left( 3 - x \right)}{x}\]
\[\lim_{x \to \pi} \frac{\sin x}{\pi - x}\]
\[\lim_{x \to a} \frac{\cos \sqrt{x} - \cos \sqrt{a}}{x - a}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \sin 2x}{1 + \cos 4x}\]
\[\lim_{x \to 2} \frac{x^2 - x - 2}{x^2 - 2x + \sin \left( x - 2 \right)}\]
\[\lim_{x \to \pi} \frac{\sqrt{2 + \cos x} - 1}{\left( \pi - x \right)^2}\]
\[\lim_{x \to 0} \left( \cos x + \sin x \right)^{1/x}\]
Write the value of \[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]
Write the value of \[\lim_{x \to 0^+} \left[ x \right] .\]
\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
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
\[\lim_{x \to \pi/4} \frac{4\sqrt{2} - \left( \cos x + \sin x \right)^5}{1 - \sin 2x}\] is equal to
\[\lim_{x \to 2} \frac{\sqrt{1 + \sqrt{2 + x} - \sqrt{3}}}{x - 2}\] 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
The value of \[\lim_{n \to \infty} \left\{ \frac{1 + 2 + 3 + . . . + n}{n + 2} - \frac{n}{2} \right\}\]
Evaluate the following limits: `lim_(x -> 2)[(x^(-3) - 2^(-3))/(x - 2)]`
if `lim_(x -> 2) (x^"n"- 2^"n")/(x - 2)` = 80 then find the value of n.
Evaluate the following limits: `lim_(x ->3) [sqrt(x + 6)/x]`
Evaluate the following limit:
`lim_(x->3)[sqrt(x+6)/x]`
Evaluate the following limit.
`lim_(x->3)[sqrt(x + 6)/x]`
Evaluate the following limit.
`lim_(x->5)[(x^3 -125)/(x^5 - 3125)]`
