Correct Answer - Option 2 : 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:
Let L = \(\rm \mathop {\lim }\limits_{x\rightarrow 0}\frac{\sin x\cdot (e^{x})}{x^{2}}\)
\(=\rm \mathop {\lim }\limits_{x\rightarrow 0}\frac{\cos xe^{x}+\sin xe^{x}}{2x}\)
\(=\rm \mathop {\lim }\limits_{x\rightarrow 0}\frac{\cos xe^{x}}{2x}+\rm \mathop {\lim }\limits_{x\rightarrow 0}\frac{\sin xe^{x}}{2x}\)
= L1 + L2
Now, L-1 = \(\rm \mathop {\lim }\limits_{x\rightarrow 0}\frac{\cos xe^{x}}{2x}\)
\(=\rm \mathop {\lim }\limits_{x\rightarrow 0}\frac{-\sin xe^{x}+\cos xe^{x}}{2}\)
Putting the limit x = 0
\(=\frac{1}{2}\)
Now, L2 = \(\rm \mathop {\lim }\limits_{x\rightarrow 0}\frac{\sin xe^{x}}{2x}\)
= \(\rm \mathop {\lim }\limits_{x\rightarrow 0}\frac{\sin x}{x} \times \frac{e^{x}}{2}\)
\(= 1 \times \frac{1}{2} = \frac 12\)
So, L = L1 + L2
= \(\frac{1}{2}+\frac{1}{2} = 1\)
Hence, option 2 is correct.