# The Limit Power Rule $ \lim_{x \to a} x^n = a^n $ ^f2joqi ## Proof We’re trying to evaluate $\lim_{x \to a} x^n = L$ which exists if given any $\epsilon$ we can come up with a range of $x$ values such that the following inequality is satisfied $|x^n - L| < \epsilon$. Recall the [[Binomial Theorem]]: $ (x+y)^n = \sum_{k=0}^n \binom n k x^{n-k}y^k $ We start by looking at $x^n$ and applying the binomial theorem to it: $ \begin{align} x^n &= (a + (x-a))^n \\ x^n &= \sum_{k=0}^n \binom n k a^{n-k}(x-a)^k && \text{peal off first element} \\ x^n &= a^n + \sum_{k=1}^n \binom n k a^{n-k}(x-a)^k && \text{subtract and apply absolute value} \\ |x^n - a^n| &= \left|\sum_{k=1}^n \binom n k a^{n-k}(x-a)^k \right| \end{align} $ Next apply the [[Absolute Value#The Triangle Inequality|triangle inequality]] $n$ times to get: $ \begin{align} |x^n - a^n| &= \left|\sum_{k=1}^n \binom n k a^{n-k}(x-a)^k \right| \\ |x^n - a^n| &\le \sum_{k=1}^n \binom n k |a^{n-k}| \cdot |x-a|^k \\ |x^n - a^n| &\le \sum_{k=1}^n \binom n k |a|^{n-k} \cdot |x-a|^k \end{align} $ Now recall that per the definition of the limit, $\lim_{x \to a}x^n = a^n$ holds true if for any given $\epsilon > 0$, we can come up with a positive value $\delta$ such that for all $x$ values within the range $a - \delta < x < a + \delta$ the following holds: $|x^n - a^n| < \epsilon$. Our range will be $a-\delta < x < a + \delta$ or $|x-a| < \delta$ so we can say that $ |x^n - a^n| < \sum_{k=1}^n \binom n k |a|^{n-k} \cdot \delta^k $ We can now apply the [[Binomial Theorem]] again to simplify the right side of the above. We start by taking $x^n = a^n + \sum_{k=1}^n \binom n k a^{n-k}(x-a)^k$ from above, and substituting $x$ and $a$ like such: $x = \delta + |a|$, and $a = |a|$. $ \begin{gather} (\delta + |a|)^n = |a|^n + \sum_{k=1}^n \binom n k |a|^{n-k}(\delta+|a|-|a|)^k \\ (\delta + |a|)^n - |a|^n = \sum_{k=1}^n \binom n k |a|^{n-k}\delta^k \\ \end{gather} $ Therefore $|x^n - a^n| < \sum_{k=1}^n \binom n k |a|^{n-k} \cdot \delta^k$ can be simplified to: $ |x^n - a^n| < (\delta + |a|)^n - |a|^n $ This gives us a way to calculate $\delta$ from $\epsilon$, proving that in fact $\lim_{x \to a}x^n = a^n$. $ \begin{gather} (\delta + |a|)^n - |a|^n = \epsilon \\ (\delta + |a|)^n = \epsilon + |a|^n \\ \delta + |a| = \sqrt[n] {\epsilon + |a|^n} \\ \delta = \sqrt[n] {\epsilon + |a|^n} - |a| \end{gather} $