Asked by Jenna
I am studying for a final exam and in our review packet we have this question, which I am having trouble with:
lim (x-->0) ((e^x)+x)^(1/x)
I tried this problem two ways.
The first was with L'Hopitals rule...
lim(x-->0) (1/x)((e^x)+x)^(1/x-1) * (e^x+1)
but that doesn't seem to help because I don't know what to do with (1/x-1)
Then I tried it like this:
y= ((e^x)+x)^(1/x)
ln y= ln((e^x)+x)^(1/x)
ln y= (1/x)ln((e^x)+x)
and then apply l'hopitals rule (because the above yields 0/0... but I always end up with an x in the denominator.
How do I solve this?
lim (x-->0) ((e^x)+x)^(1/x)
I tried this problem two ways.
The first was with L'Hopitals rule...
lim(x-->0) (1/x)((e^x)+x)^(1/x-1) * (e^x+1)
but that doesn't seem to help because I don't know what to do with (1/x-1)
Then I tried it like this:
y= ((e^x)+x)^(1/x)
ln y= ln((e^x)+x)^(1/x)
ln y= (1/x)ln((e^x)+x)
and then apply l'hopitals rule (because the above yields 0/0... but I always end up with an x in the denominator.
How do I solve this?
Answers
Answered by
bobpursley
keep going
ln y= 1/x * ln (e^x+x)
again= 1/1 * 1/(e^x+x)* (e^x+1)
and that limit is 2
so since you evaluated the limit of ln y,
then lim y must be e^2
ln y= 1/x * ln (e^x+x)
again= 1/1 * 1/(e^x+x)* (e^x+1)
and that limit is 2
so since you evaluated the limit of ln y,
then lim y must be e^2
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