Calc I: Raising Limits to Functional Exponents

[LemuelUhuru]

Homework Statement

Suppose $f(x)$ and $g(x)$ \rightarrow 0 as x \rightarrow 0+. Find examples of functions f and g with these properties and such that:

a.) $\lim_{x\rightarrow 0+} { f(x)^{g(x)} = 0 }$

Homework Equations

None

The Attempt at a Solution

Let $f(x) = 2^x-1$ and $g(x) = x$

$\displaystyle \lim_{x\rightarrow 0+} ({2^x-1})^{x} = (2^{x^2}-1^{x}) = (1-1) = 0$

My experience with limits is basic, recently while experimenting with expressions approaching infinity I was told that if you obtain an indeterminate then you need to further simplify the expression, in this case substituting x for zero immediately reduced my expression to one which is an determinate. I imagine this means there is no need to simplify and that my solution is invalid, therefore I am confused on how to proceed.

Thank you in advanced for the help.

 

[Mandelbroth]

$\displaystyle \lim_{x\rightarrow 0+} ({2^x-1})^{x} = (2^{x^2}-1^{x}) = (1-1) = 0$

Take another look at this. What did you do wrong?

 

[Ray Vickson]

Homework Statement

Suppose $f(x)$ and $g(x)$ \rightarrow 0 as x \rightarrow 0+. Find examples of functions f and g with these properties and such that:

a.) $\lim_{x\rightarrow 0+} { f(x)^{g(x)} = 0 }$

Homework Equations

None

The Attempt at a Solution

Let $f(x) = x^2-1$ and $g(x) = x$

$\displaystyle \lim_{x\rightarrow 0+} ({2^x-1})^{x} = (2^{x^2}-1^{x}) = (1-1) = 0$

My experience with limits is basic, recently while experimenting with expressions approaching infinity I was told that if you obtain an indeterminate then you need to further simplify the expression, in this case substituting x for zero immediately reduced my expression to one which is an determinate. I imagine this means there is no need to simplify and that my solution is invalid, therefore I am confused on how to proceed.

Thank you in advanced for the help.

Your "equation" $(2^x - 1)^x = 2^{x^2} - 1^x$ is false. For example, when x = 2 the left-hand-side is 9 while the right-hand-side is 15.

In general we have
(2^x-1)^x = \sum_{n=0}^{\infty} {x \choose n} (-1)^n \, 2^{(2-n)x}. The series is finite if x is a positive integer, and is infinite if x > 0 is non-integer.