#lim_(xrarr0)(e^(2x)+x)^(1//x)=# ?

2 Answers
Nov 25, 2016

Answer:

#lim_(xrarr0)(e^(2x)+x)^(1//x)=e^3#

Explanation:

#L=lim_(xrarr0)(e^(2x)+x)^(1//x)#

To undo powers like this in limits, take the natural logarithm of both sides.

#ln(L)=ln(lim_(xrarr0)(e^(2x)+x)^(1//x))#

The logarithm, being a continuous function, can be moved inside the limit.

#ln(L)=lim_(xrarr0)ln((e^(2x)+x)^(1//x))#

Move the power outside of the logarithm using log rules.

#ln(L)=lim_(xrarr0)ln(e^(2x)+x)/x#

Notice that as #xrarr0#, we have this limit in the indeterminate form #0/0#. That means L'Hospital's rule applies, and we can take the derivatives of the numerator and denominator separately in the limit.

#ln(L)=lim_(xrarr0)(d/dx(ln(e^(2x)+x)))/(d/dx(x))#

#ln(L)=lim_(xrarr0)((2e^(2x)+1)/(e^(2x)+x))/1#

#ln(L)=lim_(xrarr0)(2e^(2x)+1)/(e^(2x)+x)#

We can now evaluate the limit.

#ln(L)=(2e^0+1)/(e^0+0)#

#ln(L)=(2+1)/1#

#ln(L)=3#

#L=e^3#

Nov 25, 2016

Answer:

#e^3#

Explanation:

#(e^(2x)+x)^(1/x)= e^2(1+x/e^(2x))^(1/x)# now, making #y = x/e^(2x)#
#(e^(2x)+x)^(1/x)=e^2(1+y)^(1/(y e^(2x))) = e^2((1+y)^(1/y))^(1/(e^(2x)))#

Now #lim_(x->0) =>lim_(y(x)->0)# so

#lim_(x->0)(e^(2x)+x)^(1/x)=e^2lim_(x->0)(lim_(y->0)(1+y)^(1/y))^(1/e^(2x))=e^2 xx e=e^3#