# How do you evaluate #[ ( x )^(2x) ]# as x approaches 0+?

I don't think this can be done algebraically. I think you'll need l"Hopital's rule.

Applying the rule gets us

We conclude that

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To evaluate [ ( x )^(2x) ] as x approaches 0+, we can use the limit definition. Taking the natural logarithm of the expression, we have ln[( x )^(2x)]. Applying the properties of logarithms, this simplifies to 2x * ln(x). Now, we can evaluate the limit of this expression as x approaches 0+. Using L'Hôpital's rule, we differentiate the numerator and denominator with respect to x. The derivative of 2x is 2, and the derivative of ln(x) is 1/x. Taking the limit as x approaches 0+, we get 2 * (1/0), which is undefined. Therefore, the limit of [ ( x )^(2x) ] as x approaches 0+ does not exist.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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