Let $ f(x) = \frac{x}{\sqrt{1- x}}$ + $\frac{\sqrt{1- x}}{x}$. If $ \lim_{x _ 1^-} f(x) = l $ and $ \lim_{x \to m} f(x) = \frac{5}{2} $, then the set of all possible finite values of $ l $ and $ m $ is
The given function is: \[ f(x) = \frac{x}{\sqrt{1 - x}} + \sqrt{1 - x} \] Step 1: First, find \( \lim_{x \to 1^-} f(x) \): \[ f(x) = \frac{x}{\sqrt{1 - x}} + \sqrt{1 - x} \] As \( x \to 1^- \), both \( \frac{x}{\sqrt{1 - x}} \) and \( \sqrt{1 - x} \) behave such that the first term approaches infinity and the second approaches 0. \[ \lim_{x \to 1^-} f(x) = \infty \] Thus, \( l = 0 \).
Step 2: Now, we calculate \( \lim_{x \to m} f(x) = \frac{5}{2} \), giving the value of \( m \) to be \( \frac{2}{5} \).
Thus, the set of all possible finite values of \( l \) and \( m \) is \( \left\{ 0, \frac{2}{5}, \frac{3}{5} \right\} \).
For \( \alpha, \beta, \gamma \in \mathbb{R} \), if \[ \lim_{x \to 0} \frac{x^2 \sin(\alpha x) + (\gamma - 1)e^{x^2}}{\sin(2x - \beta x)} = 3, \] then \( \beta + \gamma - \alpha \) is equal to:
If $\lim_{x \to 1} \frac{(x-1)(6+\lambda \cos(x-1)) + \mu \sin(1-x)}{(x-1)^3} = -1$, where $\lambda, \mu \in \mathbb{R}$, then $\lambda + \mu$ is equal to