\(f(x) = \left\lfloor 1 + x \right\rfloor + \frac{\alpha^{2\left\lfloor x \right\rfloor + \left\{ x \right\}} + \left\lfloor x \right\rfloor - 1}{2\left\lfloor x \right\rfloor + \left\{ x \right\}} \)
\(\lim_{{x \to 0^-}} f(x) = \alpha - \frac{4}{3}\)
\(⇒\) \(\lim_{{x \to 0^-}} \left[ 1 + \left\lfloor x \right\rfloor + \frac{\alpha^{x + \left\lfloor x \right\rfloor} + \left\lfloor x \right\rfloor - 1}{x + \left\lfloor x \right\rfloor} \right] = \alpha - \frac{4}{3}\)
\(⇒\) \(\lim_{{h \to 0^-}} \left[ 1 - 1 + \frac{\alpha^{-h - 1} - 1 - 1}{-h - 1} \right] = \alpha - \frac{4}{3}\)
\(∴\) \(\frac{\alpha^{-1} - 2}{-1} = \alpha - \frac{4}{3}\)
\(⇒\) \(3α^2 – 10α + 3 = 0\)
\(∴\) \(α = 3 \ or\ \frac{1}{3}\)
\(∵\) α in integer, hence \(α = 3\)
\(\lim_{x \to 0} \frac{e - (1 + 2x)^{\frac{1}{2x}}}{x} \quad \text{is equal to:}\)
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
Mathematically, a limit is explained as a value that a function approaches as the input, and it produces some value. Limits are essential in calculus and mathematical analysis and are used to define derivatives, integrals, and continuity.


A derivative is referred to the instantaneous rate of change of a quantity with response to the other. It helps to look into the moment-by-moment nature of an amount. The derivative of a function is shown in the below-given formula.


Read More: Limits and Derivatives