Advertisements
Advertisements
प्रश्न
`\lim_{x \to \pi/2} \frac{e^\cos x - 1}{\cos x}`
Advertisements
उत्तर
\[\text{ If } x \to \frac{\pi}{2}, \text{ then } \cos x \to 0 . \]
\[\text{ Let } y = \cos x\]
\[ = \lim_{y \to 0} \left[ \frac{e^y - 1}{y} \right]\]
\[ = 1\]
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):}`
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)`
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{2x}\]
\[\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}\]
\[\lim_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}\]
\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x - 1}\]
\[\lim_{x \to 2} \frac{x - 2}{\sqrt{x} - \sqrt{2}}\]
\[\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}\]
\[\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x\sqrt{a^2 + ax}}\]
\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\]
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - \sqrt{x + 1}}{2 x^2}\]
\[\lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^2 - 1}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 + x}}{\sqrt{1 + x^3} - \sqrt{1 + x}}\]
\[\lim_{x \to 0} \frac{a^x + a^{- x} - 2}{x^2}\]
\[\lim_{x \to 0} \frac{a^x + b^x - 2}{x}\]
\[\lim_{x \to 0} \frac{9^x - 2 . 6^x + 4^x}{x^2}\]
\[\lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1}{x^2}\]
\[\lim_{x \to 0} \frac{a^{mx} - b^{nx}}{x}\]
\[\lim_{x \to 0} \frac{5^x + 3^x + 2^x - 3}{x}\]
\[\lim_{x \to 0} \frac{a^x + b^ x - c^x - d^x}{x}\]
\[\lim_{x \to a} \frac{\log x - \log a}{x - a}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)}\]
\[\lim_{x \to 0} \frac{\log \left| 1 + x^3 \right|}{\sin^3 x}\]
\[\lim_{x \to 0} \frac{e^x - 1}{\sqrt{1 - \cos x}}\]
\[\lim_{x \to 0} \frac{e^x - x - 1}{2}\]
`\lim_{x \to 0} \frac{e^\tan x - 1}{\tan x}`
\[\lim_{x \to 0} \frac{e^{bx} - e^{ax}}{x} \text{ where } 0 < a < b\]
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
\[\lim_{x \to \infty} \left\{ \frac{3 x^2 + 1}{4 x^2 - 1} \right\}^\frac{x^3}{1 + x}\]
\[\lim_{x \to a} \left\{ \frac{\sin x}{\sin a} \right\}^\frac{1}{x - a}\]
\[\lim_{x \to 0} \frac{\sin x}{\sqrt{1 + x} - 1} .\]
Write the value of \[\lim_{x \to \pi/2} \frac{2x - \pi}{\cos x} .\]
Evaluate: `lim_(h -> 0) (sqrt(x + h) - sqrt(x))/h`
