How do you solve this?

Question

0

672

2

\(limit( x^3*e^x, x=-inf )\)

also this

\(limit( (x^3+2x)*(cos(3/x^2))), x=0 )\)

Guest Apr 13, 2018

+33615

+3

Best Answer

When x is negative e^x goes to zero much faster than x³ goes to -infinity, so: \(\lim_{x\rightarrow -\infty}(x^3e^x)=0\)

The value of cosine is always bounded such that it's magnitude is never greater than 1.

Hence: \(\lim_{x\rightarrow 0}((x^3+2x)\cos(3/x^2))=0\)

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Alan Apr 13, 2018

0

I see, thank you very much

Guest Apr 15, 2018

Alan · Answer 1 · 2018-04-13T02:51:17

When x is negative e^x goes to zero much faster than x³ goes to -infinity, so: \(\lim_{x\rightarrow -\infty}(x^3e^x)=0\)

The value of cosine is always bounded such that it's magnitude is never greater than 1.

Hence: \(\lim_{x\rightarrow 0}((x^3+2x)\cos(3/x^2))=0\)

.