Correct Answer - Option 4 : e
-2
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
\(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)
Analysis:
\(y = \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 1}}{{x + 2}}} \right)^{2x + 1}}\)
\(y = \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 2 - 1}}{{x + 2}}} \right)^{2x + 1}}\)
\(y = \mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{1}{{\left( {x + 2} \right)}}} \right)^{2x + 1}}\)
Taking log on both sides, we get
\(\log y = \mathop {\lim }\limits_{x \to \infty } \left( {2x + 1} \right)\log \left( {1 - \frac{1}{{x + 2}}} \right)\)
\( = \mathop {\lim }\limits_{x \to \infty } \frac{{\log \left( {1 - \frac{1}{{x + 2}}} \right)}}{{\frac{1}{{\left( {2x + 1} \right)}}}}\)
\(= \frac{0}{0}\) from
Using L Hospital’s rule we get
\(\log y = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{{\left( {1 - \frac{1}{{x + 2}}} \right)}} \times \frac{1}{{{{\left( {x + 2} \right)}^2}}}}}{{ - \frac{2}{{{{\left( {2x + 1} \right)}^2}}}}}\)
\( = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{\left( {x + 2} \right)}}{{\left( {x + 1} \right)}} \times \frac{1}{{{{\left( {x + 2} \right)}^2}}}}}{{ - \frac{2}{{{{\left( {2x + 1} \right)}^2}}}}}\)
\( = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {2x + 1} \right)}^2}}}{{ - 2\left( {x + 1} \right)\left( {x + 2} \right)}}\)
\( = \mathop {\lim }\limits_{x \to \infty } - \frac{1}{2}\frac{{{{\left( {2 + \frac{1}{x}} \right)}^2}}}{{\left( {1 + \frac{1}{x}} \right)\left( {1 + \frac{2}{x}} \right)\;}}\)
\(\log y = \frac{{ - 4}}{2}\)
y = e-2
