Question #7bf46

Question

1292 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-05T23:34:22

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

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#