<s>[INST] 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 $20$.
 [/INST] 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.
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