#lim_(xrarr0)(e^(2x)+x)^(1//x)=# ?
2 Answers
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
#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#
Explanation:
Now