Evaluate the limit $\lim\limits_{x\to \infty}\left(\dfrac{20^x-1}{19x}\right)^{\frac{1}{x}}$.
My Attempt
$$\lim_{x\to \infty}\left(\frac{20^x-1}{19x}\right)^{\frac{1}{x}}=\lim_{x\to \infty}\left(\frac{(1+19)^x-1}{19x}\right)^{\frac{1}{x}}\\=\lim_{x\to \infty}\left(1+\frac{x-1}{1·2}(19)+\frac{(x-1)(x-2)}{1·2·3}(19)^2+\cdots\right)^{\frac{1}{x}}$$
After this I could not proceed. The answer given is $20$.
Best Answer
We can see that our limit is actually of the indeterminate form of ${\infty}^0$, which we can use L'hopital's on.
We will work with $y=(\frac{20^x-1}{19x})^{\frac{1}{x}}$ for now.
Taking the ln of both sides, $$\ln(y) = \frac{1}{x}\cdot \ln(\frac{20^x-1}{19x})$$
We now take $$\lim_{x\to \infty}\ln(y) = \lim_{x\to \infty}\frac{1}{x}\cdot \ln(\frac{20^x-1}{19x}) = \lim_{x\to \infty}\frac{1}{x}\cdot [\ln(20^x-1)-\ln(19x)]$$
We know $\frac{\ln(x)}{x}$ approaches 0 as x goes to infinity so our expression
$$=\lim_{x\to \infty}\frac{1}{x}\cdot \ln(20^x)=\frac{1}{x}\cdot x\cdot \ln(20) = \ln(20)$$
Just to recap, we now have $\lim_{x\to \infty} \ln(y) = \ln(20)$.
We can say that $\lim_{x\to \infty} (\frac{20^x-1}{19x})^{\frac{1}{x}} = \lim_{x\to \infty} y = \lim_{x\to \infty} e^{\ln(y)}$
By (Why is $\lim\limits_{x\to\infty} e^{\ln(y)} = e^{\,\lim\limits_{x\to\infty} \ln(y)}$?),
We can say that $\lim_{x\to \infty} e^{\ln(y)} = e^{\lim_{x\to \infty} \ln(y)} = e^{\ln(20)} = 20$