Let f(x) = lim_{n rightarrow infty} sum_{r=0} | vedclass

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(B) The expression inside the summation is $\frac{2	an 	heta}{1 - 	an^2 	heta} = 	an(2	heta)$,where $	heta = \frac{x}{2^{r+1}}$.Thus,the term i

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(B) The expression inside the summation is $\frac{2	an 	heta}{1 - 	an^2 	heta} = 	an(2	heta)$,where $	heta = \frac{x}{2^{r+1}}$.Thus,the term is $	an(2 \cdot \frac{x}{2^{r+1}}) = 	an(x/2^r)$.However,the sum given is $\sum_{r=0}^n 	an(x/2^{r+1})$. Using the identity $	an 	heta = \cot 	heta - 2\cot(2	heta)$,the sum telescopes to $f(x) = \lim_{n ightarrow \infty} (\cot(x/2^{n+1}) - \cot x) = \frac{1}{x} - \cot x$. Wait,re-evaluating the sum: $\sum_{r=0}^n 	an(x/2^{r+1}) = \frac{1}{x} - \cot x$ is incorrect. The standard identity is $	an 	heta = \cot 	heta - 2\cot 2	heta$. Actually,the sum $\sum_{r=0}^n 	an(x/2^{r+1})$ converges to $\frac{1}{x}$ as $n 	o \infty$. Thus $f(x) = \frac{1}{x} - \cot x$ is not correct. Let us use the limit $f(x) = \lim_{n 	o \infty} \sum_{r=0}^n 	an(x/2^{r+1}) = x$. Given the limit $\lim_{x 	o 0} \frac{e^x - e^{f(x)}}{x - f(x)}$,if $f(x) = x$,the expression is $0/0$. Applying $L$'Hopital's rule: $\lim_{x 	o 0} \frac{e^x - e^{f(x)}f'(x)}{1 - f'(x)}$. Since $f(x) = x$,$f'(x) = 1$,this leads to $e^0(1-1)/(1-1)$,which is indeterminate. Re-evaluating the sum: $\sum_{r=0}^n 	an(x/2^{r+1}) = \frac{1}{x} - \cot x$. As $x 	o 0$,$f(x) \approx x/3$. Using the limit $\lim_{x 	o 0} \frac{e^x - e^{f(x)}}{x - f(x)} = \lim_{u 	o 0} \frac{e^u - 1}{u} = 1$ where $u = x - f(x)$.

Let $f(x) = \lim_{n \rightarrow \infty} \sum_{r=0}^n \left( \frac{2\tan(x/2^{r+1})}{1 - \tan^2(x/2^{r+1})} \right)$. Then $\lim_{x \rightarrow 0} \frac{e^x - e^{f(x)}}{x - f(x)}$ is equal to . . . . . . .

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