# Question #9f75a

Nov 24, 2017

The limit is ${e}^{3}$.

#### Explanation:

The keys to solving this are to take the logarithm of the given function then use Taylor series centered at 0 (Maclaurin series) to determine how the expressions $\left[f \left(x\right)\right]$ and $\left\{f \left(x\right)\right\}$ can be simplified when $x$ is close to zero. Then use Taylor series again to calculate the resulting limit before exponentiating to get the final answer.

Let $f \left(x\right) = \tan \frac{x}{x}$, let $g \left(x\right) = {\left(\left[f \left(x\right)\right] + {x}^{2}\right)}^{\frac{1}{\left\{f \left(x\right)\right\}}}$, and let
$h \left(x\right) = \ln \left(g \left(x\right)\right) = \frac{\ln \left(\left[f \left(x\right)\right] + {x}^{2}\right)}{\left\{f \left(x\right)\right\}}$, where the last equality follows by a property of logarithms.

Now when $x$ is close to zero (within 1 unit of 0 is good enough in this case), then $\left[f \left(x\right)\right] = \left[\tan \frac{x}{x}\right] = 1$ since $1 < \tan \frac{x}{x} < 2$ when $- 1 < x < 1$ (also see the Taylor series expansions below).

When $- 1 < x < 1$ and $x \ne 0$ we also have $\left\{f \left(x\right)\right\} = \left\{\tan \frac{x}{x}\right\} = \tan \frac{x}{x} - 1 = {x}^{2} / 3 + \frac{2}{15} {x}^{4} + \cdots$. The reason for this follows from the Taylor expansion for $f \left(x\right) = \tan \left(x\right)$, which starts $\tan \left(x\right) = x + {x}^{3} / 3 + \frac{2}{15} {x}^{5} + \cdots$.

This implies that $\tan \frac{x}{x} = 1 + {x}^{2} / 3 + \frac{2}{15} {x}^{4} + \cdots$ for $- 1 < x < 1$ when $x \ne 0$ (actually it works for $- \frac{\pi}{2} < x < \frac{\pi}{2}$ when $x \ne 0$).

Certainly then, when $x$ is sufficiently close to zero it follows that $\left\{\tan \frac{x}{x}\right\} = {x}^{2} / 3 + \frac{2}{15} {x}^{4} + \cdots$. But this is equal to $\tan \frac{x}{x} - 1$ for $x$ sufficiently close to zero.

The Taylor series for $\ln \left(1 + {x}^{2}\right)$ is $\ln \left(1 + {x}^{2}\right) = {x}^{2} - {x}^{4} / 2 + {x}^{6} / 3 - {x}^{8} / 4 + \cdots$ for $- 1 < x < 1$.

Hence, when $x$ is sufficiently close to zero, $h \left(x\right) = \ln \frac{1 + {x}^{2}}{\tan \frac{x}{x} - 1} = \frac{{x}^{2} - {x}^{4} / 2 + \cdots}{{x}^{2} / 3 + \frac{2}{15} {x}^{4} + \cdots}$.
As $x \to 0$, the dominant terms are those of the lowest powers, so ${\lim}_{x \to 0} h \left(x\right) = \frac{1}{\frac{1}{3}} = 3$ (L'Hopital's Rule could also be used twice here).

Therefore, ${\lim}_{x \to 0} g \left(x\right) = {\lim}_{x \to 0} {e}^{h \left(x\right)} = {e}^{{\lim}_{x \to 0} h \left(x\right)} = {e}^{3}$.