Correct Answer - Option 1 :
\(\frac{1}{10}\)
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.
\(\rm\Rightarrow \frac{\mathrm{d} \sqrt{x}}{\mathrm{d} x}=\frac{1}{2\sqrt{x}}\)
Calculation:
\(\rm \mathop {\lim }\limits_{x\rightarrow 0}\frac{\sqrt{3x+9}-3}{5x}\)
Apply L-Hospital rule,
\(=\rm \mathop {\lim }\limits_{x\rightarrow 0}\frac{\frac{\mathrm{d} \sqrt{3x+9}-3}{\mathrm{d} x}}{\frac{\mathrm{d} (5x)}{\mathrm{d} x}}\)
\(=\rm \mathop {\lim }\limits_{x\rightarrow 0}\frac{3}{2\sqrt{3x+9} \times(5)}\)
Putting the limit x = 0
\(\therefore \rm \mathop {\lim }\limits_{x\rightarrow 0}\frac{\sqrt{3x+9}-3}{5x}=\frac{1}{10}\)
Hence, option 1 is correct.
