Limits by Logarithms

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When we want to evaluate a limit, it is sometimes easier to calculate the logarithm of a limit than calculating the very limit itself.

$$\lim_{x\to a} f(x)^{g(x)} = \exp \left [ \lim_{x\to a} \ln f(x)^{g(x)} \right ] = \exp \left[\lim_{x\to a} g(x) \ln f(x) \right ]$$

Prove $\displaystyle \lim_{x\to0^+} x^x = 1$.

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Solution: Let $y$ denote the value of this limit, and because the limit is in the form of $0^0$, which is an indeterminate form, then we consider taking the log of this function:

$$\ln y = \lim_{x\to 0^+} \ln (x^x) = \lim_{x\to 0^+} { x \ln x} .$$

Since the limit is in the form of $0\times \infty$, which is yet another indeterminate form, the next natural step is to consider L'Hôpital's rule:

$$\ln y = \lim_{x\to 0^+} { x \ln x} = \lim_{x\to 0^+} \dfrac{ \ln x}{1/x} \mathop{\LARGE =}^{\text{L'H}} \lim_{x\to 0^+} \dfrac{ 1/x}{-1/x^2} = \lim_{x\to 0^+} -x = 0.$$

Hence, $y = e^0 = 1$.

• L'Hôpital's Rule

• Indeterminate forms

• Limits by factoring