Limits as x approaches Infinity: e^(-x) and (1/x + 1)

[Geronomo]

Hey, I am in need of desperate help or I am dead for my test.

1) lim. x^2 e^-x as x approaches infinity = ?

2) lim. (1/x + 1) as x approaches infinity = ?

Thanks and it would be great if work was shown so i could actually understand what's going on.

 

[skeeter]

Re: Limits as x approaches Infinity

Hey, I am in need of desperate help or I am dead for my test.

lim. x^2 e^-x as x approaches infinity = ?

x²e^-x = x²/e^x ... now use L'Hopital's rule to determine the limit.


lim. (1/x + 1) as x approaches infinity = ?

if you can't figure out this limit just by looking at it, then call the undertaker.

now, get a grip and use some common sense ... what happens to the value of the fraction 1/x as x gets really large?

 

[galactus]

If you're not familiar with L'Hopital's rule, you could make a substitution.

Let \(\displaystyle \L\\u=e^{-x}\)

\(\displaystyle \L\\x^{2}=ln(u)^{2}\)

Then you have \(\displaystyle \L\\\lim_{u\to\0^{+}}[u\cdot{ln(u)^{2}}]\)

Now, it's easy to see the limit?. Just another way to go about it.