Correct Answer - Option 3 :
\(\frac{4}{3}\)
Concept:
L' hospital’s rule is used for a function who takes 0/0 or ∞/∞ form.
In this, we differentiate numerator and denominator until they take a finite value.
Calculation:
At, x → 0, the function takes a form 0/0, therefore, we can use L’ hospital’s rule
\(\mathop {{\rm{Lt}}}\limits_{x \to 0} \frac{{{{\log }_e}\left( {1 + 4x} \right)}}{{{e^{3x}} - 1}} = \frac{4}{{\left( {1 + 4x} \right)3{e^{3x}}}}\)
\( = \frac{4}{{\left( {1 + 0} \right)3{e^0}}} = \frac{4}{3}\)
