Lim _x →∞x^55^x=lim _x →∞x^5e^x log 5 lim _x →∞x^51+m x+m^2 x^22 !+… . where m=log 5 lim _x →∞11x^5+mx^4+m^22 x^3+m^36 x^2+m^424 x+m^5120+m^6 x720+… =1∞=0 A ...

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Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

limx→∞x55x=

Question

$lim x \to \infty \frac{x^{5}}{5^{x}} =$

0

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Solution

The correct option is A $0$
$lim x \to \infty \frac{x^{5}}{5^{x}} = lim x \to \infty \frac{x^{5}}{e^{x log 5}}$
$lim x \to \infty \frac{x^{5}}{1 + m x + \frac{m^{2} x^{2}}{2!} + \dots .}$ where $m = log 5$

$lim x \to \infty \frac{1}{\frac{1}{x^{5}} + \frac{m}{x^{4}} + \frac{m^{2}}{2 x^{3}} + \frac{m^{3}}{6 x^{2}} + \frac{m^{4}}{24 x} + \frac{m^{5}}{120} + \frac{m^{6} x}{720} + \dots}$

$= \frac{1}{\infty} = 0$

Alternate solution:
Applying L hosptal rule $5$ times, we observe that the numerator will have a constant term, but the denominator will have exponential term. So, limit $= 0$

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