mathop {Limit}limits_{x to infty } ,frac{{{{l | vedclass

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(B) ધારો કે $L = \mathop {Limit}\limits_{x 	o \infty } \frac{{2^{\frac{x^n}{e^x}}} - {3^{\frac{x^n}{e^x}}}}{x^n}$.કોઈપણ $n \in N$ માટે $\mathop {Limi

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(B) ધારો કે $L = \mathop {Limit}\limits_{x 	o \infty } \frac{{2^{\frac{x^n}{e^x}}} - {3^{\frac{x^n}{e^x}}}}{x^n}$.કોઈપણ $n \in N$ માટે $\mathop {Limit}\limits_{x 	o \infty } \frac{x^n}{e^x} = 0$ હોવાથી,આપણે $u = \frac{x^n}{e^x}$ લઈએ,જ્યાં $x 	o \infty$ ત્યારે $u 	o 0$.તેથી $L = \mathop {Limit}\limits_{u 	o 0} \frac{2^u - 3^u}{x^n}$.$x^n = u \cdot e^x$ હોવાથી,$L = \mathop {Limit}\limits_{x 	o \infty } \frac{2^u - 3^u}{u \cdot e^x} = \mathop {Limit}\limits_{u 	o 0} \left( \frac{2^u - 3^u}{u} ight) \cdot \mathop {Limit}\limits_{x 	o \infty } \frac{1}{e^x}$.પ્રમાણિત લક્ષ $\mathop {Limit}\limits_{u 	o 0} \frac{a^u - 1}{u} = \ln a$ નો ઉપયોગ કરતા,$\mathop {Limit}\limits_{u 	o 0} \frac{2^u - 3^u}{u} = \ln 2 - \ln 3 = \ln \left( \frac{2}{3} ight)$ મળે.જેમ $x 	o \infty$,તેમ $e^x 	o \infty$,તેથી $\frac{1}{e^x} 	o 0$.તેથી,$L = \ln \left( \frac{2}{3} ight) \cdot 0 = 0$.

$\mathop {Limit}\limits_{x \to \infty } \,\frac{{{{\left( {{2^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}}\,\, - \,\,{{\left( {{3^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}}}}{{{x^n}}}\,$ (જ્યાં $n \in N$) ની કિંમત શોધો.

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