Correct Answer - Option 2 :
\(\frac 4 3\)
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 ⇔ \(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)
Calculation:
Given:
\(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4}}{{3x - 6}}\)
As we can see, \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4}}{{3x - 6}}=\frac{0}{0}\)
So, Apply the L-Hospital rule here,
\(\mathop {\lim }\limits_{x \to 2} \frac{{{2x} }}{{3}}\)
After putting the limit we'll get:
\(\mathop {\lim }\limits_{x \to 2} \frac{{{2x} }}{{3}}=\frac{4}{3}\)
Hence the required value of the limit is \(\frac4 3\).
