Evaluate using Taylor polynomials: lim as t to 0

Evaluate using Taylor polynomials: $\displaystyle\lim_{t\to0}\left[\frac{(2-2t+t^2)e^t}{2t^3}-\frac{\sqrt{1+t^6}}{t^3}\right]$.

Step 1: Combine over a common denominator

The expression is $\dfrac{(2-2t+t^2)e^t-2\sqrt{1+t^6}}{2t^3}$; it suffices to expand the numerator to order $t^3$.

Step 2: Expand $(2-2t+t^2)e^t$ to order $t^3$

Using $e^t=1+t+\tfrac{t^2}2+\tfrac{t^3}6+o(t^3)$ and collecting by degree: \[t^0{:}\,2;\quad t^1{:}\,2t-2t=0;\quad t^2{:}\,t^2-2t^2+t^2=0;\quad t^3{:}\,\tfrac{t^3}3-t^3+t^3=\tfrac{t^3}3. \tag{1}\]

\[ (2-2t+t^2)e^t = 2+\frac{t^3}3+o(t^3). \tag{2} \]

Step 3: Expand $2\sqrt{1+t^6}$ to order $t^3$

Since $t^6=o(t^3)$ as $t\to0$: $2\sqrt{1+t^6}=2+o(t^3)$.

Step 4: Form the numerator and take the limit

\[ (2-2t+t^2)e^t-2\sqrt{1+t^6} = \frac{t^3}3+o(t^3) \;\Longrightarrow\; \frac{t^3/3+o(t^3)}{2t^3} = \frac16+\frac{o(t^3)}{2t^3} \xrightarrow[t\to0]{} \frac16. \tag{3} \]

Step 5: Verification

\[t=0.001\text{: numeric value}\approx0.16679,\ \text{approaching }1/6\approx0.16667\text{ as }t\to0\ \checkmark\]