Use series to evaluate limit

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limit as x approach infinity of $3(x^2)(e^{-2/x^2}-1)$. I don't know what series to use.

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3 Answers

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We have

$$3x^2(e^{-2/x^2}-1)=3x^2\left(1-\frac{2}{x^2}+O\left(\frac1{x^4}\right)-1\right)=-6+O\left(\frac1{x^2}\right)\xrightarrow{x\to\infty}-6$$

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Hint: Use the following series for the problem:

$$e^{f(x)} = \sum\limits_{n = 0}^{\infty} \dfrac{(f(x))^n}{n!}$$

Then, simplify the limit. What do you now have for the expression $3x^2(e^{-2/x^2} - 1)$?

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You can use the series $e^y=\sum_{n=0}^\infty{\frac{y^n}{n!}}$. With $y=-2/x^2$, we have $e^{-2/x^2}=\sum_{n=0}^\infty{\frac{-2^n}{(n!)x^{2n}}}$, so that you are taking the limit $$3\sum_{n=0}^\infty{\frac{-2^n}{(n!)x^{2n-2}}}-3x^2=3\sum_{n=1}^\infty{\frac{-2^n}{(n!)x^{2n-2}}}$$

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