\(lim _{x→7} \frac{18−[1−x]}{[x−3a]}\)
exist & \(a∈l\).
\(lim_{x→7} \frac{17−[−x]}{[x−3a]}\)
exist
\(RHL\) = \(lim_{x→7} \frac{17−[−x]}{[x−3a]}\) = \(\frac{25}{7-3a} [a ≠\frac{7}{3}]\)
\(LHL\) = \(lim_{x→7-} \frac{17−[−x]}{[x−3a]}\)
= \(\frac{24}{6-3a} [ a≠2]\)
For limit to exist
\(LHL = RHL\)
\(\frac{25}{7−3a}=\frac{24}{6−3a}\)
\(⇒\frac{25}{7−3a}=\frac{8}{2−a}\)
\(∴ a = -6\)
Hence, the correct option is (A): \(-6\)
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

Nature of compounds TeO₂ and TeH₂ is___________ and ______________respectively.
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