Evaluate using Taylor polynomials: lim as t to 0 frac 8th

Evaluate using Taylor polynomials: $\displaystyle\lim_{t\to0}\frac{\sqrt[8]{1+t}-\sqrt[8]{1-t}}t$.

Step 1: Expand each radical to first order

With $(1+u)^{1/8}=1+\tfrac u8+o(u)$: \[\sqrt[8]{1+t} = 1+\frac t8+o(t), \qquad \sqrt[8]{1-t} = 1-\frac t8+o(t). \tag{1}\]

Step 2: Subtract and divide by $t$

\[ \sqrt[8]{1+t}-\sqrt[8]{1-t} = \frac t4+o(t) \;\Longrightarrow\; \frac{\sqrt[8]{1+t}-\sqrt[8]{1-t}}t = \frac14+\frac{o(t)}t \xrightarrow[t\to0]{} \frac14. \tag{2} \]

Step 3: Verification

\[t=0.001\text{: numeric value}\approx0.2500001,\ \text{approaching }\tfrac14\text{ as }t\to0\ \checkmark\]