1. **State the problem:** Find the limits using the definition of the derivative (first principle): $$\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$ for each given function at the specified point. 2. **Recall the definition of derivative:** The limit given is the derivative of the function $f(x)$ at $x=a$: $$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$ 3. **Evaluate each limit as a derivative:** **i.** $\lim_{h \to 0} \frac{\tan^{-1}(1+h) - \frac{\pi}{4}}{h}$ is the derivative of $f(x) = \tan^{-1}(x)$ at $x=1$. The derivative formula: $$\frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2}$$ Evaluate at $x=1$: $$f'(1) = \frac{1}{1+1^2} = \frac{1}{2}$$ **ii.** $\lim_{h \to 0} \frac{\sec\left(\frac{\pi}{4} + h\right) - \sqrt{2}}{h}$ is the derivative of $f(x) = \sec(x)$ at $x=\frac{\pi}{4}$. Derivative formula: $$\frac{d}{dx} \sec(x) = \sec(x) \tan(x)$$ Evaluate at $x=\frac{\pi}{4}$: $$\sec\left(\frac{\pi}{4}\right) = \sqrt{2}, \quad \tan\left(\frac{\pi}{4}\right) = 1$$ So, $$f'\left(\frac{\pi}{4}\right) = \sqrt{2} \times 1 = \sqrt{2}$$ **iii.** $\lim_{h \to 0} \frac{2^{3+h} - 8}{h}$ is the derivative of $f(x) = 2^x$ at $x=3$. Derivative formula: $$\frac{d}{dx} 2^x = 2^x \ln(2)$$ Evaluate at $x=3$: $$f'(3) = 2^3 \ln(2) = 8 \ln(2)$$ **iv.** $\lim_{h \to 0} \frac{\ln(e + h) - 1}{h}$ is the derivative of $f(x) = \ln(x)$ at $x=e$. Derivative formula: $$\frac{d}{dx} \ln(x) = \frac{1}{x}$$ Evaluate at $x=e$: $$f'(e) = \frac{1}{e}$$ 4. **Final answers:** $$\boxed{\frac{1}{2}, \quad \sqrt{2}, \quad 8 \ln(2), \quad \frac{1}{e}}$$