Find limits, x->infty, x->0+, of 1 / [x * (e^(2/x) - 1)]

[green_tea]

What about this:

Find lim x--->? and lim x--->0+ for y=1/(x*(e^(2/x)-1))

I'm always stuck on problems with limes, I can never get them right... How do you guys do to solve them? Are there any general methods one can use to solve problems like the one above? Or is it all about simplifying and having a little luck?

I find this really difficult. I never had any trouble with math in highschool, but now I feel really stupid...

 

[soroban]

Re: another limes problem...

Hello, green_tea!

I must assume you're familiar with L'Hopital's Rule . . .

\(\displaystyle \text{Find: }\;\lim\,\frac{1}{x(e^{\frac{2}{x}}-1)}\;\;\text{ for }x \to \infty\,\text{ and }\,x \to 0^+}\)

\(\displaystyle \text{Note that: }\;\lim_{x\to\infty}f(x) \:=\:\frac{1}{\infty\cdot0}\;\;\text{ and }\;\;\lim_{x\to0^+}f(x) \:=\:\frac{1}{0\cdot\infty}\)

. . \(\displaystyle \text{Both are candidates for L'Hopital's Rule.}\)


\(\displaystyle \text{We have: }\;f(x) \;=\;\frac{x^{-1}}{e^{\frac{2}{x}} - 1}\)

\(\displaystyle \text{Apply L'Hopital: }\;\frac{-\frac{1}{x^2}}{-\frac{2}{x^2}e^{\frac{2}{x}}} \;=\;\frac{1}{2}\cdot\frac{1}{e^{\frac{2}{x}}}\)


\(\displaystyle \text{Then: }\;\lim_{x\to\infty}\left[\frac{1}{2}\cdot\frac{1}{e^{\frac{2}{x}}}\right] \;=\;\frac{1}{2}\cdot\frac{1}{1} \;=\;\frac{1}{2}\)

. \(\displaystyle \text{and: }\;\lim_{x\to0^+}\left[\frac{1}{2}\cdot\frac{1}{e^{\frac{2}{x}}}\right] \;=\;\frac{1}{2}\cdot\frac{1}{\infty} \;=\;0\)

 

[green_tea]

Re: another limes problem...

Hi! Thanx for responding so fast.. Unfortunatly, we're supposed to solve this without l'Hospitals rule. :| Do you know how to do that?