Let f(x)=n lim _r=0^n( (x/2^r+1)+ ^3(x/2^r+1)1- ^2(x/2^r+1)) Then x 0lime^x-e^f(x)(x-f(x))…

Question

Let $f(x)=\underset{n	o \infty}{lim}\sum _{r=0}^{n}(\frac{	an (x/{2}^{r+1})+{	an }^{3}(x/{2}^{r+1})}{1-{	an }^{2}(x/{2}^{r+1})})$ Then $\underset{x	o 0}{lim}\frac{{e}^{x}-{e}^{f(x)}}{(x-f(x))}$ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 1

Accepted Answer

Each term of the sum telescopes: using $\tan(2\theta)=\dfrac{2\tan\theta}{1-\tan^2\theta}$, one can show
$$\frac{\tan\theta+\tan^3\theta}{1-\tan^2\theta}=\tan\left(\frac{\theta\cdot2}{1}\right)_{\text{form}}=\tan\left(\frac{x}{2^r}\right)-\tan\left(\frac{x}{2^{r+1}}\right)\quad\text{(telescoping form)}$$

Summing from $r=0$ to $n$ and letting $n\to\infty$ (so $x/2^{n+1}\to0$):
$$f(x)=\tan x-\lim_{n\to\infty}\tan\left(\frac{x}{2^{n+1}}\right)=\tan x-0=\tan x$$

So
$$\lim_{x\to0}\frac{e^x-e^{\tan x}}{x-\tan x}=\lim_{x\to0}\frac{e^{\tan x}\left(e^{x-\tan x}-1\right)}{x-\tan x}=e^0\cdot1=1$$

Solutions · 1