# 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 $$20$$.

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 $$\lim_{x\to \infty} (\frac{20^x-1}{19x})^{\frac{1}{x}} = \lim_{x\to \infty} y = \lim_{x\to \infty} e^{\ln(y)}$$

We can say that $$\lim_{x\to \infty} e^{\ln(y)} = e^{\lim_{x\to \infty} \ln(y)} = e^{\ln(20)} = 20$$

• You may have arrived at the answer but you can't take limits piece by piece – Maverick Jul 14 '20 at 2:26
• I didn't really explain the justification of L'Hopital's. See Example 4 in tutorial.math.lamar.edu/classes/calci/lhospitalsrule.aspx – Ryan Yang Jul 14 '20 at 2:40
• Though had you gone ahead with L'Hopital rule you could have obtained the answer$\lim_{x\to \infty}\ln(y) = \lim_{x\to \infty}\frac{1}{x}\cdot \ln(\frac{20^x-1}{19x}) = \lim_{x\to \infty}\frac{1}{x}\cdot [\ln(20^x-1)-\ln(19x)]=\lim_{x\to\infty}\frac{1}{1-20^{-x}}\ln20-\frac{1}{x}=e^{\ln20}=20$. Thanks for showing the path. – Maverick Jul 14 '20 at 2:44
• @Maverick He is writing a division inside the ln as a different of two ln's upon which he distributes the $1/x$. Then he is treating the limits separately. That is ok on the condition that the separate limits exist. On a more informal note, $20^x$ is the dominant term and the only exponential term. – imranfat Jul 14 '20 at 3:37

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.

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!