Question:

<p><strong>Step 1: Determine f(x)</strong></p><p>Given: f(x) is degree 3, f(0)=1, f(1)=2, and 0 is a critical point with no local extreme.</p><p>Let f(x) = ax³ + bx² + cx + d. From f(0)=1: d=1</p><p>From f'(x)=3ax²+2bx+c and f'(0)=0: c=0</p><p>Since 0 has no local extreme (inflection point): f''(0)=0, so 2b=0, thus b=0</p><p>From f(1)=2: a(1)³=2-1=1, so a=1</p><p><strong>Therefore: f(x)=x³+1</strong></p><p><strong>Step 2: Evaluate the limit</strong></p><p>$$\lim_{x \to 0} (x^3+1)^{\frac{1}{\tan x - x}}$$</p><p>This is of form 1^∞. Take natural logarithm:</p><p>$$L = \lim_{x \to 0} \frac{\ln(x^3+1)}{\tan x - x}$$</p><p><strong>Step 3: Apply L'Hôpital's rule</strong></p><p>As x→0: ln(x³+1)→0 and tanx-x→0 (form 0/0)</p><p>$$L = \lim_{x \to 0} \frac{\frac{3x^2}{x^3+1}}{\sec^2 x - 1}$$</p><p>Using sec²x-1=tan²x and tanx≈x+x³/3:</p><p>$$L = \lim_{x \to 0} \frac{3x^2}{(\tan x)^2} = \lim_{x \to 0} \frac{3x^2}{(x+\frac{x^3}{3})^2} = \lim_{x \to 0} \frac{3x^2}{x^2(1+\frac{x^2}{9})^2} = \frac{3}{1} = 3$$</p><p><strong>∴ Answer: e³</strong></p>