1. **Problem (a):** Find all values of $a$ such that $$\lim_{x \to 1} \left( \frac{1}{x-1} - \frac{a}{x^2-1} \right)$$ exists. 2. **Rewrite the expression:** Note that $x^2 - 1 = (x-1)(x+1)$, so $$\frac{a}{x^2-1} = \frac{a}{(x-1)(x+1)}.$$ 3. **Combine the terms over a common denominator:** $$\frac{1}{x-1} - \frac{a}{(x-1)(x+1)} = \frac{(x+1)}{(x-1)(x+1)} - \frac{a}{(x-1)(x+1)} = \frac{x+1 - a}{(x-1)(x+1)}.$$ 4. **Analyze the limit:** As $x \to 1$, the denominator $(x-1)(x+1) \to 0$, so the limit will exist only if the numerator also goes to zero at $x=1$ to cancel the zero in the denominator. 5. **Set numerator zero at $x=1$:** $$1 + 1 - a = 2 - a = 0 \implies a = 2.$$ 6. **Substitute $a=2$ and simplify:** $$\lim_{x \to 1} \frac{x+1 - 2}{(x-1)(x+1)} = \lim_{x \to 1} \frac{x - 1}{(x-1)(x+1)} = \lim_{x \to 1} \frac{\cancel{x - 1}}{\cancel{(x-1)}(x+1)} = \lim_{x \to 1} \frac{1}{x+1} = \frac{1}{2}.$$ --- 7. **Problem (b):** Approximate the value of $$\lim_{x \to 0} \frac{3^x - 2^x}{x}$$ 8. **Interpretation:** This is the definition of the derivative at $x=0$ of the function $f(x) = 3^x - 2^x$. 9. **(i) Construct a table of values near $x=0$:** | $x$ | $\frac{3^x - 2^x}{x}$ | |-----|-----------------------| | 0.1 | $\frac{3^{0.1} - 2^{0.1}}{0.1} \approx \frac{1.116123 - 1.071773}{0.1} = 0.4435$ | | 0.01 | $\frac{3^{0.01} - 2^{0.01}}{0.01} \approx \frac{1.01156 - 1.00695}{0.01} = 0.461$ | | 0.001 | $\frac{3^{0.001} - 2^{0.001}}{0.001} \approx \frac{1.00116 - 1.00069}{0.001} = 0.47$ | | -0.1 | $\frac{3^{-0.1} - 2^{-0.1}}{-0.1} \approx \frac{0.896 - 0.933}{-0.1} = 0.37$ | | -0.01 | $\frac{3^{-0.01} - 2^{-0.01}}{-0.01} \approx \frac{0.9886 - 0.9931}{-0.01} = 0.45$ | From the table, values approach approximately $0.46$ to $0.47$. 10. **(ii) Confirm using graphical evidence:** The function $$y = \frac{3^x - 2^x}{x}$$ near $x=0$ approaches about $0.46$, consistent with the table. --- **Final answers:** (a) $a = 2$ and the limit is $\frac{1}{2}$. (b) The limit is approximately $0.46$.