How come the limit of $e^x$ divided by its expansion about $0$ is not equal to $1$ as $x$ goes to $∞$? I think it's it might be something about the remainder term in the expansion.
Limit of an function divided by its expansion
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limits
1 Answers
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Absolutely. When $x$ grows, the missing terms become dominant. No power of $x$ grows as fast as the exponential.
Even the ratio of successive expansions goes to infinity.
$$\frac{1+x+\dfrac{x^2}2+\dfrac{x^3}6}{1+x+\dfrac{x^2}2}\approx\frac x3.$$
$$\frac{1+x+\dfrac{x^2}2+\dfrac{x^3}6+\dfrac{x^4}{4!}}{1+x+\dfrac{x^2}2}\approx\frac{x^2}{12}.$$
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0Alright. What if x is small like (-1,1) then i think the reminder would be very small making the expansion and function pretty much equal?? – 2017-02-07
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0@Nitish: if not, the expansions woud be of strictly no use. – 2017-02-07
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0Yes. Thanks for helping me out – 2017-02-07