Question:

**Step 1: Find the Taylor expansion of $f(x) = \frac{\tan x}{x}$ near $x = 0$.** The Taylor series expansion for $\tan x$ around $x=0$ is given by: $$ \tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7) $$ Dividing by $x$, we obtain the expansion for $f(x)$: $$ f(x) = \frac{\tan x}{x} = 1 + \frac{x^2}{3} + \frac{2x^4}{15} + O(x^6) $$ **Step 2: Determine $[f(x)]$ and $\{f(x)\}$ as $x \to 0$.** As $x \to 0$, $x^2 > 0$, so $f(x) = 1 + \frac{x^2}{3} + \frac{2x^4}{15} + \dots$ implies that $f(x)$ approaches $1$ from values slightly greater than $1$ (i.e., $f(x) \to 1^+$). Therefore, the greatest integer function of $f(x)$ is: $$ [f(x)] = 1 $$ The fractional part function of $f(x)$ is: $$ \{f(x)\} = f(x) - [f(x)] = \left(1 + \frac{x^2}{3} + \frac{2x^4}{15} + O(x^6)\right) - 1 = \frac{x^2}{3} + \frac{2x^4}{15} + O(x^6) $$ **Step 3: Substitute into the limit expression.** The given limit expression is $\lim_{x \to 0}\left([f(x)] + x^2\right)^{\frac{1}{\{f(x)\}}}$. Substituting the expressions for $[f(x)]$ and $\{f(x)\}$: $$ \lim_{x \to 0}\left(1 + x^2\right)^{\frac{1}{\frac{x^2}{3} + \frac{2x^4}{15} + O(x^6)}} $$ This limit is of the indeterminate form $1^\infty$. We can evaluate it using the standard result $\lim_{u \to 0} (1+g(u))^{1/h(u)} = e^{\lim_{u \to 0} \frac{g(u)}{h(u)}}$, where $g(u) \to 0$ and $h(u) \to 0$. In this case, $g(x) = x^2$ and $h(x) = \frac{x^2}{3} + \frac{2x^4}{15} + O(x^6)$. **Step 4: Evaluate the exponent limit.** We need to find $\lim_{x \to 0} \frac{g(x)}{h(x)}$: $$ \lim_{x \to 0} \frac{x^2}{\frac{x^2}{3} + \frac{2x^4}{15} + O(x^6)} $$ Factor out $x^2$ from the denominator: $$ \lim_{x \to 0} \frac{x^2}{x^2\left(\frac{1}{3} + \frac{2x^2}{15} + O(x^4)\right)} $$ Cancel $x^2$ (since $x \ne 0$ in the limit): $$ \lim_{x \to 0} \frac{1}{\frac{1}{3} + \frac{2x^2}{15} + O(x^4)} $$ As $x \to 0$, the terms with $x^2$ and higher powers go to zero: $$ \frac{1}{\frac{1}{3} + 0 + 0} = 3 $$ **Step 5: Compute the final limit.** Using the result from Step 4, the limit is: $$ e^{\lim_{x \to 0} \frac{x^2}{\{f(x)\}}} = e^3 $$ The value of the limit is $e^3$.