Limit x approach infinity x[ ( 1 + a/x )^ 1+ 1/x - x^-1/x ( x+ a ) ]
limx→∞x[(1+ax)+1x−x−1x(x+a)]=\lim \limits_{x\to \infty} x[ ( 1 + \frac{a}{x} )+ \frac{1}{x} - \frac{x^{-1}}{x} ( x+ a ) ]=x→∞limx[(1+xa)+x1−xx−1(x+a)]=limx→∞[(x+a)+1−(1+ax)]=\lim \limits_{x\to \infty} [ ( x + a )+ 1 - ( 1+ \frac{a}{x} ) ]=x→∞lim[(x+a)+1−(1+xa)]= limx→∞[x+a−ax]=∞\lim \limits_{x\to \infty} [ x + a - \frac{a}{x} ]=\inftyx→∞lim[x+a−xa]=∞
As we know limx→∞1x=0\lim \limits_{x\to \infty} \frac{1}{x}=0x→∞limx1=0 and limx→∞x=∞\lim \limits_{x\to \infty} x=\inftyx→∞limx=∞. so the expression increases indefinitely.
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