Limit xe^(1/x) - x as x --> ∞: Confused?

[cue928]

So we've been asked to take the following limit as x--> infinity: xe^(1/x)-x
I started by dividing through by 1/x to use L'Hopital's rule, getting ((e^(1/x)/(1/x))-x/(1/x)

u = 1/x, du = -1/x^2; on the LH term, this cancels out, leaving e^1/x and on the RH term, I got 1/(-1/x^2) - this is where I'm confused; on the left term, the exponent goes to infinity, and on the RH side, I still get division by zero and it doesn't look like continuing to take the derivative is going to get me anywhere. What am I missing?

 

[hunt_mat]

Use:
<br /> e^{x}=1+x+\frac{x^{2}}{2!}+\cdots +\frac{x^{n}}{n!}+\cdots<br />

 

[hunt_mat]

Or if f(x)=xe^{\frac{1}{x}} then
<br /> f&#039;(x)=e^{\frac{1}{x}}-\frac{e^{\frac{1}{x}}}{x}<br />

 

[cue928]

Sorry but we haven't used a technique such as that. I've got to stick to using LH's rule.

 

[hunt_mat]

As you said let u=1/x and if x\rightarrow\infty then u\rightarrow 0, so the limit you need to compute is:
<br /> \lim_{u\rightarrow 0}\frac{e^{u}-1}{u}<br />
can you differentiate exp(u)-1 w.r.t. u?