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Find the limit?

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2 Answers

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.
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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.

Comments

Comment

Lim [3 ( X +h ) ^( 2( x +h) - 3 X^ (2X )] / ( X + h - x)  =
h →0
 
   3 x ^ (2x ) - 3 X ^( 2X) / h = 0/0
 
 
  3 ( X^2X) ( 1 - xh ) →3X^2X = 3
  h →0 +

Comments

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).

Comment