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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

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\[\lim_{x \rightarrow 0+} (e^x- x)^(5/x) \]
could you use L hopital's rule?
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)

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Other answers:

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

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