Correct Answer - Option 1 : 1
Concept:
L-Hospital Rule: Let f(x) and g(x) be two functions
Suppose that we have one of the following cases,
i. \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \;\frac{0}{0}\)
ii. \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \;\frac{\infty }{\infty }\)
Then we can apply L-Hospital Rule ⇔ \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \;\mathop {\lim }\limits_{x \to a} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\)
Note: We have to differentiate both the numerator and denominator with respect to x unless and until \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = l \ne \frac{0}{0}\)where l is a finite value.
Calculation:
\(\rm \mathop {\lim } \limits_{x\rightarrow 0}\frac{ln\left ( 1+3x \right )}{e^{3x}-1}\)
Using the L-Hospital rule, we get
\(= \rm \mathop {\lim}\limits_{x\rightarrow 0}\frac{\frac{\mathrm{d} \ln(1+3x)}{\mathrm{d} x}}{\frac{\mathrm{d} (e^{3x}-1)}{\mathrm{d} x}}\)
\(\rm =\lim_{x\rightarrow 0}\frac{1\times 3}{1+3x}\times \frac{1}{e^{3x}\times3}\)
Putting the value of limit
= 1
Hence, option 1 is correct.