1. **State the problem:** Find the limit $$\lim_{h \to 0} \frac{1}{h} \left( \frac{1}{\sqrt{1+h}} - 1 \right)$$ and use it to find $$f'(1)$$ for $$f(x) = \frac{1}{\sqrt{x}}$$. Also, discuss $$f(0)$$ and $$f'(0)$$. 2. **Recall the definition of the derivative:** The derivative of $$f$$ at $$x$$ is $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$. 3. **Apply the limit to find $$f'(1)$$:** Given $$f(x) = \frac{1}{\sqrt{x}}$$, then $$f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{\frac{1}{\sqrt{1+h}} - 1}{h}$$ which matches the given limit. 4. **Simplify the limit expression:** Multiply numerator and denominator inside the limit by the conjugate of the numerator: $$\frac{1}{h} \left( \frac{1}{\sqrt{1+h}} - 1 \right) \cdot \frac{\frac{1}{\sqrt{1+h}} + 1}{\frac{1}{\sqrt{1+h}} + 1} = \frac{1}{h} \cdot \frac{\left( \frac{1}{\sqrt{1+h}} \right)^2 - 1^2}{\frac{1}{\sqrt{1+h}} + 1}$$ 5. **Calculate numerator difference:** $$\left( \frac{1}{\sqrt{1+h}} \right)^2 - 1 = \frac{1}{1+h} - 1 = \frac{1 - (1+h)}{1+h} = \frac{-h}{1+h}$$ 6. **Substitute back:** $$\frac{1}{h} \cdot \frac{-h/(1+h)}{\frac{1}{\sqrt{1+h}} + 1} = \frac{1}{h} \cdot \frac{-h}{(1+h) \left( \frac{1}{\sqrt{1+h}} + 1 \right)}$$ 7. **Cancel $$h$$:** $$= \frac{\cancel{1}}{\cancel{h}} \cdot \frac{-\cancel{h}}{(1+h) \left( \frac{1}{\sqrt{1+h}} + 1 \right)} = \frac{-1}{(1+h) \left( \frac{1}{\sqrt{1+h}} + 1 \right)}$$ 8. **Take the limit as $$h \to 0$$:** $$\lim_{h \to 0} \frac{-1}{(1+h) \left( \frac{1}{\sqrt{1+h}} + 1 \right)} = \frac{-1}{1 \cdot (1 + 1)} = \frac{-1}{2}$$ 9. **Therefore,** $$f'(1) = -\frac{1}{2}$$ 10. **Discuss $$f(0)$$ and $$f'(0)$$:** Since $$f(x) = \frac{1}{\sqrt{x}}$$, as $$x \to 0^+$$, $$f(x) \to +\infty$$, so $$f(0)$$ is undefined. Also, $$f'(x) = -\frac{1}{2} x^{-3/2}$$, which tends to $$-\infty$$ as $$x \to 0^+$$, so $$f'(0)$$ is also undefined. **Final answers:** $$\lim_{h \to 0} \frac{1}{h} \left( \frac{1}{\sqrt{1+h}} - 1 \right) = -\frac{1}{2}$$ $$f'(1) = -\frac{1}{2}$$ $$f(0)$$ and $$f'(0)$$ are undefined.