# 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 \( \displaystyle \lim_{x\to0^+} x^x = 1\).

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, \(y = e^0 = 1 \).

\[ \Large \lim_{n\to\infty} \left( n \sin \dfrac \pi n\right)^{\left( 1 + \frac \pi n\right)^n} \]

The limit above has a closed form. Find the value of this closed form.

If this limit can be approximated as \(3.19\times 10^\eta\), where \(\eta \) is an integer, find \(\eta\).

You may use a calculator for the final step of your calculation.

## See Also

**Cite as:**Limits by Logarithms.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/limits-by-logarithms/