# Evaluate $\lim_{x\to 1^{+}}\left(\frac{x}{1+x}\right)^x$

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?

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

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 $$\frac{1}{2}$$).

• Yes, I got it. I think there is a typo in the book. Oct 31 '20 at 8:59

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

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

• Can't we just substitute in 1 directly? Oct 31 '20 at 9:09
• @Ameet Sharma I proved that in our case it's possible. It's not always possible. Oct 31 '20 at 9:10
• @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 $A(a)>0$. Nov 1 '20 at 6:41

$$\lim_{x\to 1^{+}}\left(\frac{x}{1+x}\right)^x=\frac12$$
$$\lim_{x\to 0^{+}}\left(\frac{x}{1+x}\right)^x=\lim_{x\to 0^{+}}\frac{x^x}{(1+x)^x}=1$$
$$\lim_{x\to \infty}\left(\frac{x}{1+x}\right)^x=\lim_{x\to \infty}\left(1-\frac{1}{1+x}\right)^x=\frac1e$$