1. **Problem Statement:** Evaluate the following limits using exponential and logarithmic concepts without L'Hôpital's rule: (a) $$\lim_{x \to \frac{\pi}{2}} (\tan x)^{\frac{\pi}{2} - x}$$ (b) $$\lim_{x \to 1^+} (x - 1)^{1 - x}$$ (c) $$\lim_{x \to 0^+} (1 + x)^{\frac{1}{x}}$$ (d) $$\lim_{x \to 0} \frac{x - \sin^{-1} x}{x^3}$$ (e) $$\lim_{x \to \frac{\pi}{2}} (\sin x)^{\tan x}$$ (f) $$\lim_{\theta \to \frac{\pi}{2}} \frac{1 - \cos \theta}{1 + \cos 2\theta}$$ (g) $$\lim_{x \to 0} \frac{\cos x - 1}{x}$$ (h) $$\lim_{x \to 0} \frac{\ln x - 1}{x - e}$$ 2. **Key formulas and rules:** - Use the fact that $$a^b = e^{b \ln a}$$ to rewrite limits involving powers. - Use series expansions or known limits for $$\sin x$$, $$\cos x$$, $$\tan x$$, $$\ln(1+x)$$ near 0. - For small $$x$$, $$\sin x \approx x - \frac{x^3}{6}$$, $$\cos x \approx 1 - \frac{x^2}{2}$$, $$\tan x \approx x$$. 3. **Step-by-step solutions:** (a) $$\lim_{x \to \frac{\pi}{2}} (\tan x)^{\frac{\pi}{2} - x}$$ Rewrite as $$e^{(\frac{\pi}{2} - x) \ln(\tan x)}$$. As $$x \to \frac{\pi}{2}$$, $$\tan x \to +\infty$$ and $$\frac{\pi}{2} - x \to 0^+$$. Set $$h = \frac{\pi}{2} - x$$, so $$h \to 0^+$$. Then $$\tan x = \tan(\frac{\pi}{2} - h) = \cot h = \frac{1}{\tan h} \approx \frac{1}{h}$$ for small $$h$$. So, $$ (\tan x)^{\frac{\pi}{2} - x} = (\cot h)^h = \left(\frac{1}{h}\right)^h = e^{h \ln(1/h)} = e^{h(-\ln h)} = e^{-h \ln h} $$ As $$h \to 0^+$$, $$h \ln h \to 0$$, so $$-h \ln h \to 0$$. Therefore, limit is $$e^0 = 1$$. (b) $$\lim_{x \to 1^+} (x - 1)^{1 - x}$$ Set $$t = x - 1$$, so $$t \to 0^+$$. Rewrite as $$t^{1 - (t+1)} = t^{-t} = e^{-t \ln t}$$. As $$t \to 0^+$$, $$t \ln t \to 0$$, so $$-t \ln t \to 0$$. Limit is $$e^0 = 1$$. (c) $$\lim_{x \to 0^+} (1 + x)^{\frac{1}{x}}$$ Rewrite as $$e^{\frac{1}{x} \ln(1+x)}$$. Use expansion $$\ln(1+x) \approx x - \frac{x^2}{2}$$ for small $$x$$. So, $$ \frac{1}{x} \ln(1+x) \approx \frac{1}{x} \left(x - \frac{x^2}{2}\right) = 1 - \frac{x}{2} $$ As $$x \to 0^+$$, this tends to 1. Limit is $$e^1 = e$$. (d) $$\lim_{x \to 0} \frac{x - \sin^{-1} x}{x^3}$$ Use expansion $$\sin^{-1} x = x + \frac{x^3}{6} + O(x^5)$$. So numerator: $$ x - \sin^{-1} x = x - \left(x + \frac{x^3}{6} + ...\right) = -\frac{x^3}{6} + ... $$ Divide by $$x^3$$: $$ \frac{-\frac{x^3}{6} + ...}{x^3} = -\frac{1}{6} + ... $$ Limit is $$-\frac{1}{6}$$. (e) $$\lim_{x \to \frac{\pi}{2}} (\sin x)^{\tan x}$$ Rewrite as $$e^{\tan x \ln(\sin x)}$$. As $$x \to \frac{\pi}{2}$$, $$\sin x \to 1$$ and $$\tan x \to +\infty$$. Set $$h = \frac{\pi}{2} - x$$, so $$h \to 0$$. Then $$\sin x = \sin(\frac{\pi}{2} - h) = \cos h \approx 1 - \frac{h^2}{2}$$. So $$\ln(\sin x) \approx \ln\left(1 - \frac{h^2}{2}\right) \approx -\frac{h^2}{2}$$. Also, $$\tan x = \tan(\frac{\pi}{2} - h) = \cot h \approx \frac{1}{h}$$. Therefore, $$ \tan x \ln(\sin x) \approx \frac{1}{h} \left(-\frac{h^2}{2}\right) = -\frac{h}{2} \to 0 $$ as $$h \to 0$$. Limit is $$e^0 = 1$$. (f) $$\lim_{\theta \to \frac{\pi}{2}} \frac{1 - \cos \theta}{1 + \cos 2\theta}$$ Use $$\cos \theta \approx 0$$ at $$\theta = \frac{\pi}{2}$$. Also, $$\cos 2\theta = \cos(\pi) = -1$$. Evaluate numerator: $$1 - \cos \frac{\pi}{2} = 1 - 0 = 1$$ Denominator: $$1 + \cos \pi = 1 - 1 = 0$$ Check behavior near $$\theta = \frac{\pi}{2}$$. Set $$h = \theta - \frac{\pi}{2}$$, small. Then $$\cos \theta = \cos(\frac{\pi}{2} + h) = -\sin h \approx -h$$. So numerator: $$1 - (-h) = 1 + h$$. Also, $$\cos 2\theta = \cos(\pi + 2h) = -\cos 2h \approx - (1 - 2h^2) = -1 + 2h^2$$. Denominator: $$1 + (-1 + 2h^2) = 2h^2$$. Therefore, $$ \frac{1 + h}{2h^2} $$ As $$h \to 0$$, numerator $$\to 1$$, denominator $$\to 0^+$$. Limit tends to $$+\infty$$. (g) $$\lim_{x \to 0} \frac{\cos x - 1}{x}$$ Use expansion $$\cos x = 1 - \frac{x^2}{2} + ...$$ Numerator: $$\cos x - 1 \approx -\frac{x^2}{2}$$ Divide by $$x$$: $$\frac{-\frac{x^2}{2}}{x} = -\frac{x}{2} \to 0$$ Limit is 0. (h) $$\lim_{x \to 0} \frac{\ln x - 1}{x - e}$$ Note: As $$x \to 0$$, $$\ln x \to -\infty$$, denominator $$x - e \to -e$$. Numerator tends to $$-\infty$$, denominator tends to $$-e$$. So limit tends to $$\frac{-\infty}{-e} = +\infty$$. 4. **Final answers:** (a) 1 (b) 1 (c) e (d) $$-\frac{1}{6}$$ (e) 1 (f) $$+\infty$$ (limit does not exist finite) (g) 0 (h) $$+\infty$$ (limit diverges)