Advertisements
Advertisements
प्रश्न
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
उत्तर
\[ = \lim_{x \to a} \left\{ 1 + \frac{\sin x}{\sin a} - 1 \right\}^\frac{1}{\left( x - a \right)} \]
\[ = \lim_{x \to a} \left\{ 1 + \frac{\sin x - \sin a}{\sin a} \right\}^\frac{1}{\left( x - a \right)} \]
\[ = e {}^\lim_{x \to a} \left( \frac{\sin x - \sin a}{\sin a} \right) \times \frac{1}{\left( x - a \right)} \]
\[ = e {}^\lim_{x \to a} \left( \frac{2 \cos \left( \frac{x + a}{2} \right) \sin \left( \frac{x - a}{2} \right)}{\sin a \times \left( \frac{x - a}{2} \right) \times 2} \right) \]
\[ = e^\frac{\cos a}{\sin a} \]
` = e^\cot a`
APPEARS IN
संबंधित प्रश्न
Evaluate `lim_(x -> 0) f(x)` where `f(x) = { (|x|/x, x != 0),(0, x = 0):}`
If the function f(x) satisfies `lim_(x -> 1) (f(x) - 2)/(x^2 - 1) = pi`, evaluate `lim_(x -> 1) f(x)`.
\[\lim_{x \to 0} \frac{2x}{\sqrt{a + x} - \sqrt{a - x}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x - 1}\]
\[\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}\]
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - \sqrt{x + 1}}{2 x^2}\]
\[\lim_{x \to 4} \frac{2 - \sqrt{x}}{4 - x}\]
\[\lim_{x \to a} \frac{x - a}{\sqrt{x} - \sqrt{a}}\]
\[\lim_{x \to \sqrt{10}} \frac{\sqrt{7 + 2x} - \left( \sqrt{5} + \sqrt{2} \right)}{x^2 - 10}\]
\[\lim_{x \to \sqrt{2}} \frac{\sqrt{3 + 2x} - \left( \sqrt{2} + 1 \right)}{x^2 - 2}\]
\[\lim_{x \to 0} \frac{9^x - 2 . 6^x + 4^x}{x^2}\]
\[\lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1}{x^2}\]
\[\lim_{x \to 2} \frac{x - 2}{\log_a \left( x - 1 \right)}\]
\[\lim_{x \to \infty} \left( a^{1/x} - 1 \right)x\]
\[\lim_{x \to 0} \frac{a^x + b^ x - c^x - d^x}{x}\]
\[\lim_{x \to 0} \frac{e^x - 1 + \sin x}{x}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log \left( a - x \right)}{x}\]
\[\lim_{x \to 0} \frac{x\left( 2^x - 1 \right)}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{e^x - 1}{\sqrt{1 - \cos x}}\]
\[\lim_{x \to 5} \frac{e^x - e^5}{x - 5}\]
\[\lim_{x \to 0} \frac{e^{3 + x} - \sin x - e^3}{x}\]
\[\lim_{x \to 0} \frac{e^x - x - 1}{2}\]
\[\lim_{x \to 0} \frac{e^{bx} - e^{ax}}{x} \text{ where } 0 < a < b\]
`\lim_{x \to 0} \frac{e^x - e^\sin x}{x - \sin x}`
\[\lim_{x \to 0} \frac{a^x - a^{- x}}{x}\]
\[\lim_{x \to \pi/2} \frac{2^{- \cos x} - 1}{x\left( x - \frac{\pi}{2} \right)}\]
\[\lim_{x \to \infty} \left\{ \frac{x^2 + 2x + 3}{2 x^2 + x + 5} \right\}^\frac{3x - 2}{3x + 2}\]
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
\[\lim_{x \to 0} \frac{\sin x}{\sqrt{1 + x} - 1} .\]
Write the value of \[\lim_{x \to \pi/2} \frac{2x - \pi}{\cos x} .\]
Write the value of \[\lim_{n \to \infty} \frac{1 + 2 + 3 + . . . + n}{n^2} .\]
