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

By (Why is $\lim\limits_{x\to\infty} e^{\ln(y)} = e^{\,\lim\limits_{x\to\infty} \ln(y)}$?),

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

  • 2
    $\begingroup$ You may have arrived at the answer but you can't take limits piece by piece $\endgroup$ – Maverick Jul 14 '20 at 2:26
  • $\begingroup$ I didn't really explain the justification of L'Hopital's. See Example 4 in tutorial.math.lamar.edu/classes/calci/lhospitalsrule.aspx $\endgroup$ – Ryan Yang Jul 14 '20 at 2:40
  • $\begingroup$ 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. $\endgroup$ – Maverick Jul 14 '20 at 2:44
  • 3
    $\begingroup$ @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. $\endgroup$ – 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!


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