Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
have you tried rewriting without negative exponent ?
\[\lim_{x \rightarrow \infty}x^9e ^{-x^8}\]= \[\frac{ x^9 }{ e ^{x^8} }\] =\[\lim_{x \rightarrow \infty}\frac{ 9x^7 }{ 8e ^{x^8} }\]
thats what I have so far
that should make it obvious what the limit is... but if not I guess you could l'hospital what about (9+1) times and if that 9 l'hospital version of the limit doesn't convince you that the limit is ____ then I don't know what will :p
eventually you will get a constant on top and you will have an exponential still on bottom
so would I have to find the derivative around 7-8 more times until its just a constant at top?
\[(e^{x^8})'=(x^8)' \cdot e^{x^8}=8x^7 \cdot e^{x^8}\] I lied you won't need that many derivatives
but you will still wind up with a constant over a function going to infinity still after [,e .[re derivatove
after one more derivative *
@Zenmo you should have 9x^8 up top after using L' Hospitals rule and `-8x^7*e^(-x^8)` in the bottom. Then simplify
negative won't be there though
because we put in bottom
sorry, `8x^7e^(x^8)` in the bottom
yes true
Just want to say that you could work it using Lhopital rule just one time
but ofcourse it requires a clever but elementary trick...
[,e .[re derivatove =one more derivative lol my keyboard was skewed to the right when I was typing
Here is a neat way oldrin had shown @ganeshie8 can recall to \(\color{blue}{\text{Originally Posted by}}\) @oldrin.bataku \[\frac{x^n}{e^x}=n^n\left(\frac{x/n}{e^{x/n}}\right)^n\] so consider passing the limit in: \[\lim_{x\to\infty} n^n\left(\frac{x/n}{e^{x/n}}\right)^n=n^n\left(\lim_{u\to\infty}\frac{u}{e^u}\right)^n\] with \(u=x/n\) \(\color{blue}{\text{End of Quote}}\)
lol you have an elephant memory
Hehe =P
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