Question

$\underset{n\to \infty }{lim}(1+x)(1+x^2)(1+x^4)...(1+x^{2n}),|x|<1$, is equal to

• A. $\frac{1}{x-1}$
• B. $\frac{1}{1-x}$
• C. $1-x$
• D. $1$

Answer 1

The correct option is B $\frac{1}{1-x}$

$\underset{n\to \infty }{lim}(1+x)(1+x^2)(1+x^4)...(1+x^{2n})$

$=\frac{1}{1-x}\underset{n\to \infty }{lim}(1-x)(1+x)(1+x^2)(1+x^4)...(1+x^{2n})$

$=\frac{1}{1-x}\underset{n\to \infty }{lim}(1-x^{4n})=\frac{1}{1-x}$ since $|x|<1,\{{lim}_{n\to \infty }{x}^{4n}=0\}$