Mht Cet
Limits Continuity And Differentiability
MHT CET 2026 15th April Evening Shift
MCQ+2 / -02026
The value of f(0) so that the function $f(x) = \dfrac{(256 - 8x)^{\frac{1}{4}} - 4}{16 - 4(64 + 3x)^{\frac{1}{3}}}$, $x \neq 0$ is continuous at $x = 0$, is
