Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}\]
Advertisements
उत्तर
\[\lim_{x \to 0} \left[ \frac{\sin \left( ax \right) + bx}{ax + \sin \left( bx \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\frac{\sin ax}{ax} \times ax + bx}{ax + \frac{\sin bx}{bx} \times bx} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( \frac{\sin ax}{ax} \times a + b \right)x}{\left( a + \frac{\sin bx}{bx} \times b \right)x} \right]\]
\[ = \frac{1 \times a + b}{a + b}\]
\[ = 1\]
APPEARS IN
संबंधित प्रश्न
\[\lim_{x \to 0} \frac{x^{2/3} - 9}{x - 27}\]
\[\lim_{x \to 1} \left( \frac{1}{x^2 + x - 2} - \frac{x}{x^3 - 1} \right)\]
\[\lim_{x \to 2} \left[ \frac{1}{x - 2} - \frac{2\left( 2x - 3 \right)}{x^3 - 3 x^2 + 2x} \right]\]
\[\lim_{x \to 1} \left\{ \frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3 x^2 + 2x} \right\}\]
\[\lim_{x \to 4} \frac{x^3 - 64}{x^2 - 16}\]
\[\lim_{x \to \infty} \frac{3 x^3 - 4 x^2 + 6x - 1}{2 x^3 + x^2 - 5x + 7}\]
\[\lim_{x \to \infty} \frac{5 x^3 - 6}{\sqrt{9 + 4 x^6}}\]
\[\lim_{n \to \infty} \left[ \frac{1^2 + 2^2 + . . . + n^2}{n^3} \right]\]
\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\]
\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number.
\[\lim_{x \to - \infty} \left( \sqrt{x^2 - 8x} + x \right)\]
\[\lim_\theta \to 0 \frac{\sin 3\theta}{\tan 2\theta}\]
\[\lim_\theta \to 0 \frac{1 - \cos 4\theta}{1 - \cos 6\theta}\]
\[\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}\]
\[\lim_{x \to 0} \frac{\sin 3x + 7x}{4x + \sin 2x}\]
Evaluate the following limits:
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - 1}\]
If \[\lim_{x \to 0} kx cosec x = \lim_{x \to 0} x cosec kx,\]
\[\lim_{x \to \frac{\pi}{2}} \frac{\cot x}{\frac{\pi}{2} - x}\]
\[\lim_{x \to a} \frac{\cos x - \cos a}{\sqrt{x} - \sqrt{a}}\]
\[\lim_{x \to 1} \frac{1 - x^2}{\sin 2\pi x}\]
\[\lim_{x \to \pi} \frac{1 + \cos x}{\tan^2 x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( 4x - \pi \right)^2}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\cos x - \sin x}{\left( \frac{\pi}{4} - x \right) \left( \cos x + \sin x \right)}\]
\[\lim_{x \to 0} \frac{\log \left( 3 + x \right) - \log \left( 3 - x \right)}{x}\]
Write the value of \[\lim_{x \to 1^-} x - \left[ x \right] .\]
\[\lim_{x \to \infty} \frac{\sin x}{x}\] equals
\[\lim_{x \to 3} \frac{x - 3}{\left| x - 3 \right|},\] is equal to
\[\lim 2_{h \to 0} \left\{ \frac{\sqrt{3} \sin \left( \pi/6 + h \right) - \cos \left( \pi/6 + h \right)}{\sqrt{3} h \left( \sqrt{3} \cos h - \sin h \right)} \right\}\]
\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]
\[\lim_{n \to \infty} \frac{1 - 2 + 3 - 4 + 5 - 6 + . . . . + \left( 2n - 1 \right) - 2n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}\] is equal to
If α is a repeated root of ax2 + bx + c = 0, then \[\lim_{x \to \alpha} \frac{\tan \left( a x^2 + bx + c \right)}{\left( x - \alpha \right)^2}\]
The value of \[\lim_{x \to \infty} \frac{n!}{\left( n + 1 \right)! - n!}\]
Evaluate the following limits: if `lim_(x -> 5)[(x^"k" - 5^"k")/(x - 5)]` = 500, find all possible values of k.
If the value of `lim_(x -> 1) (1 - (1 - x))^"m"/x` is 99, then n = ______.
Evaluate the following limits: `lim_(x ->3) [sqrt(x + 6)/x]`
Evaluate the Following limit:
`lim_(x->5) [(x^3 -125)/(x^5-3125)]`
Evaluate the following limit :
`lim_(x->5)[(x^3-125)/(x^5-3125)]`
Evaluate the following limit:
`lim_(x->5)[(x^3-125)/(x^5-3125)]`
