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# How do you solve this?

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

also this

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

Guest Apr 13, 2018

#1
+27061
+2

When x is negative ex 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:   $$\lim_{x\rightarrow 0}((x^3+2x)\cos(3/x^2))=0$$

.

Alan  Apr 13, 2018
#1
+27061
+2

When x is negative ex 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:   $$\lim_{x\rightarrow 0}((x^3+2x)\cos(3/x^2))=0$$

.

Alan  Apr 13, 2018
#2
0

I see, thank you very much

Guest Apr 15, 2018