Advertisements
Advertisements
प्रश्न
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
Advertisements
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
Let a1, a2,..., an be fixed real numbers and define a function f ( x) = ( x − a1 ) ( x − a2 )...( x − an ).
What is `lim_(x -> a_1) f(x)` ? For some a ≠ a1, a2, ..., an, compute `lim_(x -> a) f(x)`
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?
If the function f(x) satisfies `lim_(x -> 1) (f(x) - 2)/(x^2 - 1) = pi`, evaluate `lim_(x -> 1) f(x)`.
if `f(x) = { (mx^2 + n, x < 0),(nx + m, 0<= x <= 1),(nx^3 + m, x > 1):}`
For what integers m and n does `lim_(x-> 0) f(x)` and `lim_(x -> 1) f(x)` exist?
\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - 1}{x}\]
\[\lim_{x \to 0} \frac{2x}{\sqrt{a + x} - \sqrt{a - x}}\]
\[\lim_{x \to 2} \frac{\sqrt{3 - x} - 1}{2 - x}\]
\[\lim_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}\]
\[\lim_{x \to 2} \frac{\sqrt{x^2 + 1} - \sqrt{5}}{x - 2}\]
\[\lim_{x \to 2} \frac{x - 2}{\sqrt{x} - \sqrt{2}}\]
\[\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x\sqrt{a^2 + ax}}\]
\[\lim_{x \to 1} \frac{\left( 2x - 3 \right) \left( \sqrt{x} - 1 \right)}{3 x^2 + 3x - 6}\]
\[\lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}, x \neq 0\]
\[\lim_{x \to \sqrt{2}} \frac{\sqrt{3 + 2x} - \left( \sqrt{2} + 1 \right)}{x^2 - 2}\]
\[\lim_{x \to 0} \frac{\log \left( 1 + x \right)}{3^x - 1}\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{x}\]
\[\lim_{x \to 0} \frac{a^x + b^x + c^x - 3}{x}\]
\[\lim_{x \to \infty} \left( a^{1/x} - 1 \right)x\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{\sin kx}\]
\[\lim_{x \to 0} \frac{a^x + b^ x - c^x - d^x}{x}\]
\[\lim_{x \to 0} \frac{e^x - 1 + \sin x}{x}\]
\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\]
\[\lim_{x \to 0} \frac{e\sin x - 1}{x}\]
\[\lim_{x \to 0} \frac{\log \left( a + x \right) - \log \left( a - x \right)}{x}\]
\[\lim_{x \to 0} \frac{\log \left( 2 + x \right) + \log 0 . 5}{x}\]
\[\lim_{x \to 0} \frac{x\left( 2^x - 1 \right)}{1 - \cos x}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)}\]
\[\lim_{x \to 0} \frac{\log \left| 1 + x^3 \right|}{\sin^3 x}\]
\[\lim_{x \to 0} \frac{e^x - 1}{\sqrt{1 - \cos x}}\]
`\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^\tan x - 1}{\tan x}`
`\lim_{x \to 0} \frac{e^\tan x - 1}{x}`
\[\lim_{x \to 0} \frac{a^x - a^{- x}}{x}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
