Find the limit of 3x^(2x) as x approaches to 0+.
That's not right, guys. Have you tried to graph it? With GeoGebra it looks like part of a parabola opening up with vertex at (0.36788, 1.43743).
If you put a point on the function and slide it toward 0 from the right it disappears at x = 0; so there’s a hole there. If you move it very close to 0 you get (0.00172952010749865, 2.93472308532316).
So the limit of 3x^(2x) as x approaches 0+ (from the right) is approximately 2.93472308532316.
I found that we need to convert to logarithms and manipulate them into an indeterminate form in order to use L'Hopital's Rule.
3x^(2x) = 3 e^(ln(x^(2x)))
= 3 e^(2x ln(x))
= 3 e^(ln(x)/(1/(2x)))
x –> 0+, ln(x) –> -inf, and 1/(2x) –> +inf, so we can use L'Hopital's Rule.
x –> 0+, ln(x)/(1/(2x)) –> (1/x)/(–1/(2x^2))
–> (–2x^2/x) –> –2x; so
x –> 0+, 3 e^(ln(x)/(1/(2x))) –> 3 e^(-2x) = 3.