Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 0} \frac{\log \left( 1 + x \right)}{3^x - 1}\]
Advertisements
उत्तर
\[\lim_{x \to 0} \left[ \frac{\log \left( 1 + x \right)}{3^x - 1} \right]\]
Dividing the numerator and the denominator by x:
\[\lim_{x \to 0} \left[ \frac{\log \left( 1 + x \right)}{x \cdot \left( \frac{3^x - 1}{x} \right)} \right] \left[ \because \lim_{x \to 0} \frac{\log \left( 1 + x \right)}{x} = 1 \lim_{x \to 0} \left( \frac{a^x - 1}{x} \right) = \log a \right]\]
\[ = \frac{1}{\log 3}\]
APPEARS IN
संबंधित प्रश्न
Find `lim_(x -> 0)` f(x) and `lim_(x -> 1)` f(x) where f(x) = `{(2x + 3, x <= 0),(3(x+1), x > 0):}`
Find `lim_(x -> 1)` f(x), where `f(x) = {(x^2 -1, x <= 1), (-x^2 -1, x > 1):}`
Evaluate `lim_(x -> 0) f(x)` where `f(x) = { (|x|/x, x != 0),(0, x = 0):}`
If f(x) = `{(|x| + 1,x < 0), (0, x = 0),(|x| -1, x > 0):}`
For what value (s) of a does `lim_(x -> a)` f(x) exists?
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{2x}\]
\[\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}\]
\[\lim_{x \to 2} \frac{\sqrt{x^2 + 1} - \sqrt{5}}{x - 2}\]
\[\lim_{x \to 2} \frac{\sqrt{1 + 4x} - \sqrt{5 + 2x}}{x - 2}\]
\[\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}, x \neq 0\]
\[\lim_{x \to \sqrt{10}} \frac{\sqrt{7 + 2x} - \left( \sqrt{5} + \sqrt{2} \right)}{x^2 - 10}\]
\[\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6}\]
\[\lim_{x \to 0} \frac{5^x - 1}{\sqrt{4 + x} - 2}\]
\[\lim_{x \to 0} \frac{a^{mx} - 1}{b^{nx} - 1}, n \neq 0\]
\[\lim_{x \to 0} \frac{9^x - 2 . 6^x + 4^x}{x^2}\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{x}\]
\[\lim_{x \to 2} \frac{x - 2}{\log_a \left( x - 1 \right)}\]
\[\lim_{x \to 0} \frac{5^x + 3^x + 2^x - 3}{x}\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{\sin kx}\]
\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\]
\[\lim_{x \to 0} \frac{e\sin x - 1}{x}\]
\[\lim_{x \to 0} \frac{e^{2x} - e^x}{\sin 2x}\]
\[\lim_{x \to a} \frac{\log x - \log a}{x - a}\]
\[\lim_{x \to 0} \frac{\log \left( 2 + x \right) + \log 0 . 5}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)}\]
\[\lim_{x \to 0} \frac{e^x - 1}{\sqrt{1 - \cos x}}\]
\[\lim_{x \to 5} \frac{e^x - e^5}{x - 5}\]
\[\lim_{x \to 0} \frac{e^{x + 2} - e^2}{x}\]
`\lim_{x \to \pi/2} \frac{e^\cos x - 1}{\cos x}`
\[\lim_{x \to 0} \frac{e^{3x} - e^{2x}}{x}\]
`\lim_{x \to 0} \frac{e^x - e^\sin x}{x - \sin x}`
\[\lim_{x \to \pi/2} \frac{2^{- \cos x} - 1}{x\left( x - \frac{\pi}{2} \right)}\]
\[\lim_{x \to 1} \left\{ \frac{x^3 + 2 x^2 + x + 1}{x^2 + 2x + 3} \right\}^\frac{1 - \cos \left( x - 1 \right)}{\left( x - 1 \right)^2}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
Evaluate: `lim_(x -> 2) (x^2 - 4)/(sqrt(3x - 2) - sqrt(x + 2))`
