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$\lim\limits_{x\to \infty}\large\frac{x^n}{e^x}$$=0$ for

$\begin{array}{1 1}(a)\;n=0\;only&(b)\;n\;is\;any\;whole\;number\\(c)\;n=2\;only&(d)\;No\;value\;of\;n\end{array}$

1 Answer

$\lim\limits_{x\to \infty}\large\frac{x^n}{e^x}\big(\frac{\infty}{\infty}\big)$
$\Rightarrow \lim\limits_{x\to\infty}\large\frac{nx^{n-1}}{e^x}$
$\Rightarrow \lim\limits_{x\to \infty}\large\frac{n!}{e^{\Large x}}$$=0$
Where n is any whole number as n! is defined for the integers and 0.
Hence (b) is the correct answer.
answered Dec 30, 2013 by sreemathi.v
 

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