$$\log L = \lim_{x \to \infty} \frac{\log \left( \int_0^x 2019^t - 2001^t dt\right)}{x} \\ \log L = \lim_{x \to \infty} \frac{2019^x-2001^x}{\int_0^x 2019^t - 2001^t dt} \\$$
I was solving some problems on the internet, and I come up with these lines on the answer. I tried to understand how this change can be possible, but I can't do that with my own strength. Please help me to go forward...
In the numerator if we insert the limit inside the $\log \left( \int_0^x 2019^t - 2001^t dt\right)$ then we see that as $x\to \infty$ the value $\left( \int_0^x 2019^t - 2001^t dt\right) \to \infty$ because $f(t)= 2019^t - 2001^t \gt 0\\$ for all $ t \in [0,\infty)$ and the function is getting integrated over $[0,\infty]$ hence the area under the curve tends to $\infty$. Therefore the limit becomes a $\frac{\infty}{\infty}$ form. Then use L'Hospital and you'll get the required next line
$\log L = \lim_{x \to \infty} \frac{2019^x-2001^x}{\int_0^x 2019^t - 2001^t dt} \\$
By the way if you have a problem differentiating the numerator then use this : $ \frac{d}{dx} \int_0^x\{f(t)-g(t)\}dt = f(x)-g(x) $