Is there anything wrong with this proof?

Question

$\lim_{n \to \infty} \int_{0}^{1} \dfrac{x^n}{1+x }dx=\lim_{n\to \infty} \xi^n \int_{0}^{1} \dfrac{dx}{1+x}=\lim_{n \to \infty} \xi ^n \ln{2}=(\ln{2}) \lim_{n \to \infty} \xi ^n =0 \qquad (0 \le \xi \le 1) $

this proof is a part of pondering in my textbook ,and i can't find fault in it ! i'm still not sure !

Answer 1

As already commented, the first equality is really weird, but the statement is true by the Lebesgue dominated convergence theorem. Observe that for $n\geqslant 1$, your integrand is easily seen to be bounded by 1 (actually it is bounded by $\frac{1}{2}$) which gives an integrable upper bound for your sequence of functions. You can hence interchange the limit and the integration and obtain zero as result.

EDIT: I guess the proof is meant along these lines:

Observe that $$\begin{aligned} \int_0^1\frac{x^n}{1+x}\mathrm dx & = \int_0^\xi\frac{x^n}{1+x}\mathrm dx + \int_\xi^1\frac{x^n}{1+x}\mathrm dx\\ & \leqslant \xi^n \int_0^\xi\frac{1}{1+x}\mathrm dx + (1-\xi) \end{aligned}$$ With this you'll get the statement in your spirit...

Answer 2

The $\xi$ in the formula is not independent of $n$ - it DOES depend on $n$. Hence $\xi$ should be replaced by $\xi_n$.

Although $\xi_n\in (0,1),$ we can not deduce that $\xi_n^n\to 0$. For example, if $$ \xi_n=1-\frac{1}{n}, $$ then $$ \xi_n^n\to\frac{1}{\mathrm{e}}. $$