JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Continuity and Differentiability

Continuity at $0$ requires the three values to agree. The left limit is ${\lim }_{x\to 0^{-}}(1+ax{)}^{1/x}=e^a$, and $f(0)=1+b$. For the right limit ${\lim }_{x\to 0^{+}}\frac{(x+4{)}^{1/2}-2}{(x+c{)}^{1/3}-2}$ to be finite the denominator must vanish at $0$, so $c^{1/3}=2$, i.e. $c=8$; the form is then $\frac{0}{0}$, and by L'Hopital it equals $\frac{\frac{1}{2\sqrt{4}}}{\frac{1}{3}\cdot 8^{-2/3}}=\frac{1/4}{1/12}=3$. Matching, $e^a=1+b=3$, so $b=2$ and $e^a=3$. Hence $e^{a}bc=3\cdot 2\cdot 8=48$.