Limit under Evaluation

Calculus Level 3

$\displaystyle \lim_{n \to \infty} \left( (1+x)(1+x^2)(1+x^4)....(1+x^{2^n}) \right)=?$

provided that $|x|<1$ .

Try for some more interesting problems of Limits and Derivatives.

\dfrac{x^2}{1-x}

\dfrac{x}{1-x}

None of the given. Doesn't exist.

\dfrac{1}{1-x}

2 solutions

Tanishq Varshney
Apr 10, 2015

Just two hints will do it.

1) Multiply and divide by $(1-x)$ and apply identity of $a^2-b^2$

2) $\displaystyle \lim_{n\to \infty} 1-x^{2n+1}=1$ as $-1<x<1$

Curtis Clement
Apr 11, 2015

$lim_{n \rightarrow\infty } \left( (1+x)(1+x^2)(1+x^4)...(1+x^{2^n}) \right)$ $= lim_{n \rightarrow\infty} (\frac{1- x^{2^{n+1}}}{1-x} ) = \frac{1}{1-x}$ Due to the fact that $|x| < 1 \iff lim_{n \rightarrow\infty} x^{2^n} = 0$

@Curtis Clement , there's a minor typo in your solution.

It should be $\large x^{2^{n+1}}$ in the numerator instead of $x^{2n}$ because the "before last step" of the "chain reaction" that occurs in the numerator is,

$\large (1-x^{2n})(1+x^{2n})=1-(x^{2n})^2=1-x^{2^{n+1}}$

Prasun Biswas - 6 years, 1 month ago

Your prays have been answered! :)

Curtis Clement - 6 years, 1 month ago

Limit under Evaluation

Try for some more interesting problems of Limits and Derivatives.

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