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2.22 Exponential growth

(iv) For $x>0$ , all the terms in the series

\exp(x)=1+x+{{x^{2}}\over{2!}}+{{x^{3}}\over{3!}}+{{x^{4}}\over{4!}}+\dots

are positive, so

\exp(x)\geq 1+x;

hence $\exp(x)\rightarrow\infty$ as $x\rightarrow\infty$ . Now let $u=-x$ , so $u\rightarrow-\infty$ as $x\rightarrow\infty$ and

\exp u=\exp(-x)=1/\exp(x)\rightarrow 0.