That's too much!

$\begin{aligned} L & = \lim_{n \to \infty} \left(2009^{2010n}+2010^{2009n}\right)^\frac 1n \\ & = \lim_{n \to \infty} \exp \left(\ln \left(2009^{2010n}+2010^{2009n}\right)^\frac 1n \right) & \small \blue{\text{where }\exp (x) = e^x} \\ & = \exp \left(\lim_{n \to \infty} \frac {\ln \left(2009^{2010n}+2010^{2009n}\right)}n \right) \\ & = \exp \left(\lim_{n \to \infty} \frac {\ln \left(2009^{2010n}\left(1+\frac {2010^{2009n}}{2009^{2010n}}\right)\right)}n \right) \\ & = \exp \left(\lim_{n \to \infty} \frac {\ln \left(2009^{2010n}\right)}n + \lim_{n \to \infty} \frac {\ln \left(1+\frac {2010^{2009n}}{2009^{2010n}}\right)}n \right) \\ & = \exp \left(\lim_{n \to \infty} \frac {2010n\ln \left(2009\right)}n + 0 \right) \\ & = \exp \left(2010\ln \left(2009\right) \right) \\ & = 2009^{2010} \end{aligned}$

Therefore $|a-b| = |2009-2010| = \boxed 1$ .

Consider the function $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ defined as $f(x) = x^{\frac{1}{x}}$ . By calculating the first derivative of $f$ we see that this function is strictly decreasing for $x>e$ , this implies $f(2009)>f(2010)$ and thus $2009^{2010} > 2010^{2009}$ . This result will be helpfull for us in evaluating the given limit. For convinience, let $a = 2009^{2010}$ and $b = 2010^{2009}$ , we know $a > b$ . Now consider the limit $\lim_{x \to \infty} \left( 1+c^x \right)^{\frac{1}{x}} = e^{\lim \frac{\ln{(1+c^x)}}{x}}$ , assuming $c >1$ , we have $\frac{\infty}{\infty}$ form, using LH rule we get $\lim_{x \to \infty} \left( 1+c^x \right)^{\frac{1}{x}} = c$ . Now we are ready to solve our given limit. $\lim_{n \to \infty} \left( a^n+b^n\right)^{1/n}$ $= b\lim_{n \to \infty} \left( 1+(\frac{a}{b})^n\right)^{1/n} = a$ . Thus the answer is $2009^{2010}$ .