1
$\begingroup$

I tried: $$\left(\frac{x}{1+x}\right)^x=A$$ $$\ln(A)=x\ln\left(\frac{x}{1+x}\right)$$ $$\lim_{x\to 1^{+}}\frac{\ln x-\ln (x+1)}{\frac1x}=\lim_{x\to 1^{+}}\frac{\frac1x-\frac1{x+1}}{\frac{-1}{x^2}}=\frac{-1}{2}$$ I applied L'Hopital rule in last step. I get $\lim(A)_{x\to 1^{+}}=e^{\frac{-1}{2}}$ But the final answer provided in the book is $e^{-1}$. where I made a mistake?

$\endgroup$
2
  • 2
    $\begingroup$ You can just substitute in 1. $(\frac{1}{1+1})^1 = (\frac{1}{2})^1 = \frac{1}{2}$ $\endgroup$ Oct 31 '20 at 8:59
  • $\begingroup$ A typo in the book, probably this limit was intended. $\endgroup$ Nov 1 '20 at 6:34
2
$\begingroup$

Hint

You cannot apply l'Hopital here since it's not indeterminated. Nevertheless, $$\lim_{x\to 1^+}\frac{\ln(x)-\ln(1+x)}{1/x}=-\ln(2).$$ I let you conclude (and the limit won't be $e^{-1}$ but $\frac{1}{2}$).

$\endgroup$
1
  • 1
    $\begingroup$ Yes, I got it. I think there is a typo in the book. $\endgroup$
    – Soheil
    Oct 31 '20 at 8:59
1
$\begingroup$

If our function is wrote in this form, you can't use l'Hopital.

Since, $e^x$ and $\ln$ are continuous functions we obtain: $$\lim_{x\rightarrow1}\left(\frac{x}{1+x}\right)^x=\lim_{x\rightarrow1}e^{x\ln\frac{x}{1+x}}=e^{\lim\limits_{x\rightarrow1}x\ln\frac{x}{1+x}}=e^{1\cdot\ln\lim\limits_{x\rightarrow1}\frac{x}{1+x}}=e^{\ln\frac{1}{2}}=\frac{1}{2}.$$

$\endgroup$
3
  • $\begingroup$ Can't we just substitute in 1 directly? $\endgroup$ Oct 31 '20 at 9:09
  • $\begingroup$ @Ameet Sharma I proved that in our case it's possible. It's not always possible. $\endgroup$ Oct 31 '20 at 9:10
  • 1
    $\begingroup$ @AmeetSharma Yes you can substitute $x=1$ because there is nothing in sight that would disturb the continuity of this function. The general result is that $A(x)^{B(x)}$ is continuous at $x=a$, if $A(x)$ and $B(x)$ are, and $A(a)>0$. $\endgroup$ Nov 1 '20 at 6:41
1
$\begingroup$

As already noticed by continuity

$$\lim_{x\to 1^{+}}\left(\frac{x}{1+x}\right)^x=\frac12$$

it seems that, fixed the typo, the original question is referring to

$$\lim_{x\to 0^{+}}\left(\frac{x}{1+x}\right)^x=\lim_{x\to 0^{+}}\frac{x^x}{(1+x)^x}=1$$

or to

$$\lim_{x\to \infty}\left(\frac{x}{1+x}\right)^x=\lim_{x\to \infty}\left(1-\frac{1}{1+x}\right)^x=\frac1e$$

note that in any case we don't need l'Hospital's rule to obtain the result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.