Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
उत्तर
\[\lim_{x \to 0} \left[ \frac{e^x + e^{- x} - 2}{x^2} \right]^\left( \frac{1}{x^2} \right) \]
\[ = \lim_{x \to 0} \left[ 1 + \frac{e^x + e^{- x} - 2}{x^2} - 1 \right]^\left( \frac{1}{x^2} \right) \]
\[ = e {}^\lim_{x \to 0} \left( \frac{e^x + e^{- x} - 2}{x^2} - 1 \right) \times \left( \frac{1}{x^2} \right) \]
\[ \because e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + . . . . . . \propto \]
\[ e^{- x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + . . . . . . \propto \]
\[ \Rightarrow e^x + e^{- x} = 2 + \frac{2 x^2}{2!} + \frac{2 x^4}{4!} + . . . . . \propto \]
\[ = e^\lim_{x \to 0} \left( \frac{2 + \frac{2 x^2}{2!} + \frac{2 x^4}{4!} . . . \propto - 2}{x^2} - 1 \right) \times \left( \frac{1}{x^2} \right) \]
\[ = e^\lim_{x \to 0} \left( \frac{\frac{2 x^2}{2!} + \frac{2 x^4}{4!} + . . . . . . \propto}{x^4} - \frac{1}{x^2} \right) \]
\[ = e^\lim_{x \to 0} \left( \frac{x^2 + \frac{x^4}{12} + . . . . . \propto - x^2}{x^4} \right) \]
\[ = e^\frac{1}{12}\]
APPEARS IN
संबंधित प्रश्न
Evaluate `lim_(x -> 0) f(x)` where `f(x) = { (|x|/x, x != 0),(0, x = 0):}`
\[\lim_{x \to 0} \frac{\sqrt{a^2 + x^2} - a}{x^2}\]
\[\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}\]
\[\lim_{x \to 1} \frac{x - 1}{\sqrt{x^2 + 3 - 2}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^2 - 1}\]
\[\lim_{x \to 2} \frac{\sqrt{x^2 + 1} - \sqrt{5}}{x - 2}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - \sqrt{x + 1}}{2 x^2}\]
\[\lim_{x \to 0} \frac{\sqrt{2 - x} - \sqrt{2 + x}}{x}\]
\[\lim_{x \to 1} \frac{\left( 2x - 3 \right) \left( \sqrt{x} - 1 \right)}{3 x^2 + 3x - 6}\]
\[\lim_{x \to \sqrt{10}} \frac{\sqrt{7 + 2x} - \left( \sqrt{5} + \sqrt{2} \right)}{x^2 - 10}\]
\[\lim_{x \to \sqrt{6}} \frac{\sqrt{5 + 2x} - \left( \sqrt{3} + \sqrt{2} \right)}{x^2 - 6}\]
\[\lim_{x \to \sqrt{2}} \frac{\sqrt{3 + 2x} - \left( \sqrt{2} + 1 \right)}{x^2 - 2}\]
\[\lim_{x \to 0} \frac{5^x - 1}{\sqrt{4 + x} - 2}\]
\[\lim_{x \to 0} \frac{a^x + a^{- x} - 2}{x^2}\]
\[\lim_{x \to 0} \frac{a^{mx} - 1}{b^{nx} - 1}, n \neq 0\]
\[\lim_{x \to 0} \frac{a^x + b^x - 2}{x}\]
\[\lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1}{x^2}\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{x}\]
\[\lim_{x \to \infty} \left( a^{1/x} - 1 \right)x\]
\[\lim_{x \to 0} \frac{a^x + b^ x - c^x - d^x}{x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)}\]
\[\lim_{x \to 0} \frac{\log \left| 1 + x^3 \right|}{\sin^3 x}\]
`\lim_{x \to \pi/2} \frac{a^\cot x - a^\cos x}{\cot x - \cos x}`
\[\lim_{x \to 0} \frac{e^{x + 2} - e^2}{x}\]
\[\lim_{x \to 0} \frac{e^x - x - 1}{2}\]
\[\lim_{x \to 0} \frac{e^{3x} - e^{2x}}{x}\]
`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`
\[\lim_{x \to 0} \frac{x\left( e^x - 1 \right)}{1 - \cos x}\]
\[\lim_{x \to 1} \left\{ \frac{x^3 + 2 x^2 + x + 1}{x^2 + 2x + 3} \right\}^\frac{1 - \cos \left( x - 1 \right)}{\left( x - 1 \right)^2}\]
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]
Write the value of \[\lim_{x \to \pi/2} \frac{2x - \pi}{\cos x} .\]
Evaluate: `lim_(x -> 2) (x^2 - 4)/(sqrt(3x - 2) - sqrt(x + 2))`
