1. **Problem a:** Find $$\lim_{x \to 0} \frac{\sin x}{x}$$ using Bernoulli-l’Hôpital’s rule. Since direct substitution gives $$\frac{0}{0}$$, apply l’Hôpital’s rule: $$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = \cos 0 = 1.$$ 2. **Problem b:** Find $$\lim_{x \to 1} \frac{e^{2(x-1)} - x^2}{(x^2 - 1)^2}$$. Direct substitution yields $$\frac{e^0 - 1}{0} = \frac{0}{0}$$ indeterminate form. Apply l’Hôpital’s rule: Differentiate numerator: $$\frac{d}{dx} (e^{2(x-1)} - x^2) = 2e^{2(x-1)} - 2x.$$ Differentiate denominator: $$\frac{d}{dx} (x^2 - 1)^2 = 2(x^2 - 1) \cdot 2x = 4x(x^2 - 1).$$ Evaluate limit: $$\lim_{x \to 1} \frac{2e^{2(x-1)} - 2x}{4x(x^2 - 1)} = \frac{2e^0 - 2}{4 \cdot 1 \cdot 0} = \frac{0}{0}$$ still indeterminate. Apply l’Hôpital’s rule again: Second derivative numerator: $$\frac{d}{dx} (2e^{2(x-1)} - 2x) = 4e^{2(x-1)} - 2.$$ Second derivative denominator: $$\frac{d}{dx} (4x(x^2 - 1)) = 4(x^2 - 1) + 4x \cdot 2x = 4(x^2 - 1) + 8x^2 = 12x^2 - 4.$$ Evaluate limit at $$x=1$$: $$\frac{4e^0 - 2}{12(1)^2 - 4} = \frac{4 - 2}{12 - 4} = \frac{2}{8} = \frac{1}{4}.$$ 3. **Problem c:** Find $$\lim_{x \to \infty} \left(x^2\right)^{\frac{1}{x}}$$. Rewrite the expression: $$\left(x^2\right)^{\frac{1}{x}} = e^{\frac{1}{x} \ln(x^2)} = e^{\frac{2 \ln x}{x}}.$$ Evaluate the exponent limit: $$\lim_{x \to \infty} \frac{2 \ln x}{x}.$$ Apply l’Hôpital’s rule since $$\frac{\infty}{\infty}$$: $$\lim_{x \to \infty} \frac{2 \ln x}{x} = \lim_{x \to \infty} \frac{2 \cdot \frac{1}{x}}{1} = \lim_{x \to \infty} \frac{2}{x} = 0.$$ Therefore, $$\lim_{x \to \infty} \left(x^2\right)^{\frac{1}{x}} = e^0 = 1.$$ 4. **Problem d:** Find $$\lim_{x \to 0^+} x^x$$. Rewrite using exponentials: $$x^x = e^{x \ln x}.$$ Evaluate the exponent limit: $$\lim_{x \to 0^+} x \ln x.$$ Rewrite as: $$\lim_{x \to 0^+} \frac{\ln x}{1/x}.$$ Apply l’Hôpital’s rule (since $$\ln x \to -\infty$$ and $$1/x \to \infty$$): Derivative numerator: $$\frac{1}{x}.$$ Derivative denominator: $$-\frac{1}{x^2}.$$ Limit becomes: $$\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0.$$ Therefore, $$\lim_{x \to 0^+} x^x = e^0 = 1.$$