1. **State the problem:** We have a piecewise function $$f(x) = \begin{cases} \frac{\sqrt{1+x} - 1}{\sqrt[3]{1+x} - 1} & x \neq 1 \\ A & x = 1 \end{cases}$$ We want to find the value(s) of $A$ such that $f(x)$ is continuous at $x=1$. 2. **Recall the continuity condition:** A function $f$ is continuous at $x=1$ if $$\lim_{x \to 1} f(x) = f(1) = A.$$ So we need to compute $$\lim_{x \to 1} \frac{\sqrt{1+x} - 1}{\sqrt[3]{1+x} - 1}$$ and set it equal to $A$. 3. **Evaluate the limit:** Let $t = 1 + x$. Then as $x \to 1$, $t \to 2$. Rewrite the limit as $$\lim_{t \to 2} \frac{\sqrt{t} - 1}{\sqrt[3]{t} - 1}.$$ 4. **Direct substitution:** Substitute $t=2$: $$\frac{\sqrt{2} - 1}{\sqrt[3]{2} - 1}.$$ This is a finite number, so the limit exists and equals this value. 5. **Therefore, the function is continuous at $x=1$ if** $$A = \frac{\sqrt{2} - 1}{\sqrt[3]{2} - 1}.$$ 6. **Summary:** The function $f$ is continuous at $x=1$ if and only if $$A = \frac{\sqrt{2} - 1}{\sqrt[3]{2} - 1}.$$