Averi Mitchell

2022-06-25

$\underset{x\to \mathrm{\infty}}{lim}\left({x}^{6}[{e}^{-\frac{1}{2{x}^{3}}}-\mathrm{cos}\left(\frac{1}{x\sqrt{x}}\right)]\right)$

Any tip on how to calculate it?

Any tip on how to calculate it?

ejigaboo8y

Beginner2022-06-26Added 29 answers

In such cases, I always find that transforming the beast into a limit at zero is easier. Perform the substitution

${x}^{3/2}=1/t$

so ${x}^{3}=1/{t}^{2}$ and ${x}^{6}=1/{t}^{4}$. Now the limit becomes

$\underset{t\to {0}^{+}}{lim}\frac{{e}^{-{t}^{2}/2}-\mathrm{cos}t}{{t}^{4}}$

and the Taylor expansion of the numerator up to degree 4 is

$1-\frac{{t}^{2}/2}{1!}+\frac{{t}^{4}/4}{2!}-1+\frac{{t}^{2}}{2!}-\frac{{t}^{4}}{4!}+o({t}^{4})=\frac{{t}^{4}}{12}+o({t}^{4})$

${x}^{3/2}=1/t$

so ${x}^{3}=1/{t}^{2}$ and ${x}^{6}=1/{t}^{4}$. Now the limit becomes

$\underset{t\to {0}^{+}}{lim}\frac{{e}^{-{t}^{2}/2}-\mathrm{cos}t}{{t}^{4}}$

and the Taylor expansion of the numerator up to degree 4 is

$1-\frac{{t}^{2}/2}{1!}+\frac{{t}^{4}/4}{2!}-1+\frac{{t}^{2}}{2!}-\frac{{t}^{4}}{4!}+o({t}^{4})=\frac{{t}^{4}}{12}+o({t}^{4})$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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