## anonymous 5 years ago I have a problem, lim as [x->0+] ((e^x)+x)^(5/x)...I have tried to just plug in 0 for all x's and I got (1)^(5/0). is there any algebraic manipulation that I can do to the ((e^x)+x)^(5/x) to get rid of that (5/x) exponent

1. anonymous

$\lim_{x \rightarrow 0+} (e^x- x)^(5/x)$

2. anonymous

could you use L hopital's rule?

3. anonymous

I was thinking of that but the only way i know how to apply L'H is when i have it in f(x)/g(x)

4. anonymous

applying L hopital's rule $\[\lim_{x \rightarrow 0+}(5/x) (e^x -x)^(5/x) [(e^x-1)/(e^x-x) + (1/x)(\ln(e^x-x))]$\] we know that 5/x when x approach to 0 = positive infinite and the positive infinite multiply by any number will be infinitive...

5. anonymous

this part is missing int he end of the lim $+ (1/x)(\ln(e^x-x)$

6. anonymous

Can't you seperate the limit? Like this? $\lim_{x \rightarrow 0+}5/x * \lim_{x \rightarrow 0+} (e^x-x)$ I feel like that's a much easier solution than trying to take it all at once since the first one approaches infinity and the 2nd = 1.