Evaluating $\lim\limits_{x\to \infty}\left(\frac{20^x-1}{19x}\right)^{\frac{1}{x}}$ 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 $20$.
 A: 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 $\lim_{x\to \infty} e^{\ln(y)} =  e^{\lim_{x\to \infty} \ln(y)} = e^{\ln(20)} = 20$
A: You can calculate the limit directly using the standard limits $\lim_{x\to \infty} x^{\frac 1x}= 1$ and $\lim_{x\to \infty} a^{\frac 1x}= 1$  for any $a>0$ as follows:
Note that
$$\frac{20}{(19x)^{\frac 1x}} =\frac{(20^x)^{\frac 1x}}{(19x)^{\frac 1x}} > \frac{(20^x-1)^{\frac 1x}}{(19x)^{\frac 1x}} >\frac{(20^x-\frac 12\cdot 20^x)^{\frac 1x}}{(19x)^{\frac 1x}}  = \frac{20\cdot \left(\frac 12\right)^{\frac 1x}}{(19x)^{\frac 1x}} $$
Now, squeezing gives the desired result.
A: All right, we have to find the limit of $\left(\frac{20^x -1}{19x} \right)^{1/x}$ as $x$ goes to infinity, let's call the expression $\left(\frac{20^x -1}{19x} \right)^{1/x}$ $f(x)$ . We should observe that as $x$ grows bigger and bigger $20^x -1$ will be very close to $20^x$ and hence we can approximate it by that. Using this approximation we have
$$
\lim_{x \to \infty} \left(\frac{20^x -1}{19x} \right)^{1/x} = \lim_{x\to \infty} \left(\frac{20^x}{19x} \right)^{1/x} \\
\lim_{x \to \infty} f(x) = \lim_{x\to \infty}\frac{20}{ (19x)^{1/x} }
$$
Now, let's have look at the limit of $(19x)^{-1/x}$,
$$
(19x)^{-1/x} =y \\
ln (y) = -\frac{ln(19)}{x} - \frac{ln (x)}{x} \\
\lim_{x\to \infty} ln(y) = 0 - \lim_{x \to \infty} \frac{ln (x)}{x} $$
We can prove that $\lim_{x \to \infty} \frac{ln (x)}{x}=0$
$$\text{Hence}, ~  \lim_{x\to \infty} ln(y)=0 \implies  \lim_{x\to \infty}(19x)^{-1/x} = 1
$$
Putting this value in our original limit calculation we have
$$
\lim_{x \to \infty} f(x) = 20 
$$
Hope it helps!
