Correct Answer - Option 3 : 0
Concept:
\(\rm \mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) + g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) + \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)\)
\(\rm \mathop {\lim }\limits_{x\; \to \;0} \frac { (e^x - 1) }{x} = \log e = 1\)
log mn = n log m
Calculation:-
\(\rm \displaystyle\lim_{x \rightarrow 0} \frac{e^x + e^{-x}-2}{\sin x}\) is \(\left ( \frac{0}{0} \right )\) form ,
Apply L - hospital rule, differentiate numerator and denominator w.r.t x .
= \(\rm \lim_{x\to0} \frac{e^{x}- e^{-x}+0 }{\cos x}\)
Put a limit of x in the above eq.
= \(\rm \frac{e^{0}- e^{-0}+0 }{\cos 0}= \frac {1-1}{1}\)
= 0
The correct option is 1.
