Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log a}{x}\]
Advertisements
उत्तर
\[\lim_{x \to 0} \left[ \frac{\log \left( a + x \right) - \log a}{x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\log \left( \frac{a + x}{a} \right)}{x} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\log \left( 1 + \frac{x}{a} \right)}{\frac{x}{a} \times a} \right]\]
\[ = \frac{1}{a} \times 1\]
\[ = \frac{1}{a}\]
APPEARS IN
संबंधित प्रश्न
Find `lim_(x -> 5) f(x)`, where f(x) = |x| - 5
\[\lim_{x \to 0} \frac{x^{2/3} - 9}{x - 27}\]
\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\]
\[\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^2 + 4\sqrt{3}x - 15}\]
\[\lim_{x \to 1} \left( \frac{1}{x^2 + x - 2} - \frac{x}{x^3 - 1} \right)\]
\[\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2}\]
\[\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{4}{x^3 - 2 x^2} \right)\]
\[\lim_{x \to 27} \frac{\left( x^{1/3} + 3 \right) \left( x^{1/3} - 3 \right)}{x - 27}\]
\[\lim_{x \to \infty} \frac{x}{\sqrt{4 x^2 + 1} - 1}\]
\[\lim_{x \to \infty} \left[ \left\{ \sqrt{x + 1} - \sqrt{x} \right\} \sqrt{x + 2} \right]\]
\[\lim_{x \to 0} \left[ \frac{x^2}{\sin x^2} \right]\]
\[\lim_{x \to 0} \frac{\sin x \cos x}{3x}\]
\[\lim_{x \to 0} \frac{\sin 5x}{\tan 3x}\]
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - \cos dx}\]
\[\lim_\theta \to 0 \frac{\sin 3\theta}{\tan 2\theta}\]
\[\lim_{x \to 0} \frac{\tan 3x - 2x}{3x - \sin^2 x}\]
\[\lim_{x \to 0} \frac{\tan x - \sin x}{\sin 3x - 3 \sin x}\]
\[\lim_{x \to 0} \frac{3 \sin^2 x - 2 \sin x^2}{3 x^2}\]
\[\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\]
Evaluate the following limits:
\[\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - 1}\]
Evaluate the following limit:
\[\lim_{h \to 0} \frac{\left( a + h \right)^2 \sin\left( a + h \right) - a^2 \sin a}{h}\]
\[\lim_{x \to \pi} \frac{\sin x}{\pi - x}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{\cos x} - \sqrt{\sin x}}{x - \frac{\pi}{4}}\]
\[\lim_{x \to a} \frac{\cos x - \cos a}{\sqrt{x} - \sqrt{a}}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \sin 2x}{1 + \cos 4x}\]
\[\lim_{n \to \infty} \frac{\sin \left( \frac{a}{2^n} \right)}{\sin \left( \frac{b}{2^n} \right)}\]
\[\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{2} - \cos x - \sin x}{\left( 4x - \pi \right)^2}\]
\[\lim_{x \to 0} \frac{\log \left( 3 + x \right) - \log \left( 3 - x \right)}{x}\]
Write the value of \[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
\[\lim_{x \to 0} \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]
\[\lim_{x \to a} \frac{x^n - a^n}{x - a}\] is equal at
\[\lim 2_{h \to 0} \left\{ \frac{\sqrt{3} \sin \left( \pi/6 + h \right) - \cos \left( \pi/6 + h \right)}{\sqrt{3} h \left( \sqrt{3} \cos h - \sin h \right)} \right\}\]
If \[\lim_{x \to 1} \frac{x + x^2 + x^3 + . . . + x^n - n}{x - 1} = 5050\] then n equal
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\] is equal to
\[\lim_{x \to \pi/3} \frac{\sin \left( \frac{\pi}{3} - x \right)}{2 \cos x - 1}\] is equal to
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 -> 3) [sqrt(x + 6)/x]`
