Question #7bf46

1 Answer
Feb 5, 2017

lim_(x->0) (e^x+30x)^(1/x) =e^31

Explanation:

Consider the function:

f(x) = (e^x+30x)^(1/x)

and take its logarithm:

ln f(x) = ln((e^x+30x)^(1/x))

using the properties of logarithm this is:

ln f(x) = 1/x ln(e^x+30x)

Now evaluate:

lim_(x->0) ln f(x) = lim_(x->0) ln(e^x+30x)/x

it is in the indeterminate form 0/0 so we can use l'Hospital's rule:

lim_(x->0) ln f(x) = lim_(x->0) (d/dx ln(e^x+30x))/(d/dx x) = lim_(x->0) (e^x+30)/(e^x+30x) = 31

Since lnx is a continuous function in (0,+oo):

31 = lim_(x->0) ln f(x) = ln(lim_(x->0) f(x))

so that:

lim_(x->0) f(x) =e^31