Correct Answer - Option 3 : 6/7
Concept:
L-Hospital Rule:
Let f(x) and g(x) be two functions
Suppose that we have one of the following cases,
I. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}\)
II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{\infty }{\infty }\)
Then we can apply L-Hospital Rule as:
\(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)
Calculation:
\(\Rightarrow \rm \lim_{x\rightarrow 2}\frac{2x^2\;-\;2x \;-\;4}{3x^2\;-\;5x\;-\;2} = \frac{0}{0}\)
So, by using L'hospital's rule i.e. Differentiating numerator and denominator w.r.t. x.
⇒ \(\rm \lim_{x\rightarrow 2}\frac{2x^2\;-\;2x \;-\;4}{3x^2\;-\;5x\;-\;2} = \rm \lim_{x\rightarrow 2}\frac{4x\;-\;2 }{6x\;-\;5}\)
\(= \rm \frac{4\;\times\;2\;-\;2 }{6\;\times\;2\;-\;5} = \frac{6}{7}\) .
∴ Option 3 is the correct answer.
