Question
The value of $$\mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {{2^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}} - {{\left( {{3^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}}}}{{{x^n}}}$$ (where $$n\, \in \,N$$ ) is :
A.
$$\log \,n\left( {\frac{2}{3}} \right)$$
B.
$$0$$
C.
$$n\,\log \,n\left( {\frac{2}{3}} \right)$$
D.
not defined
Answer :
$$0$$
Solution :
$$\eqalign{
& L = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( {{2^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}} - {{\left( {{3^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}}}}{{{x^n}}} \cr
& \,\,\,\,\, = \mathop {\lim }\limits_{x \to \infty } \frac{{{{\left( 3 \right)}^{\frac{{{x^n}}}{{{e^x}}}}}\left( {{{\left( {\frac{2}{3}} \right)}^{\frac{{{x^n}}}{{{e^x}}}}} - 1} \right)}}{{{x^n}}} \cr
& {\text{Now, }}\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{n!}}{{{e^x}}} = 0 \cr
& \left( {{\text{Applying L'Hospital rule }}n{\text{ times}}} \right) \cr
& {\text{Hence,}} \cr
& L = \mathop {\lim }\limits_{x \to \infty } {\left( 3 \right)^{\frac{{{x^n}}}{{{e^x}}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {{{\left( {\frac{2}{3}} \right)}^{\frac{{{x^n}}}{{{e^x}}}}} - 1} \right)}}{{\frac{{{x^n}}}{{{e^x}}}}}\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{e^x}}} \cr
& \,\,\,\,\, = 1 \times \log \left( {\frac{2}{3}} \right) \times 0 \cr
& \,\,\,\,\, = 0 \cr} $$