Here is the step-by-step solution for the provided problems:

a) $\lim_{x \to 2} \left( \frac{1}{x-2} - \frac{4}{x^2-4} \right)$

Step 1: Factorize $x^2 - 4$

$x^2 - 4 = (x-2)(x+2)$

Rewrite the second term: $\frac{4}{x^2-4} = \frac{4}{(x-2)(x+2)}$

So the expression becomes: $\lim_{x \to 2} \left( \frac{1}{x-2} - \frac{4}{(x-2)(x+2)} \right)$

Step 2: Combine terms under a common denominator

The common denominator is $(x-2)(x+2)$: $\frac{1}{x-2} - \frac{4}{(x-2)(x+2)} = \frac{(x+2) - 4}{(x-2)(x+2)} = \frac{x-2}{(x-2)(x+2)}$

Simplify: $\frac{x-2}{(x-2)(x+2)} = \frac{1}{x+2}, \quad \text{for } x \neq 2.$

Step 3: Evaluate the limit as $x \to 2$

Substitute $x = 2$ into $\frac{1}{x+2}$: $\lim_{x \to 2} \frac{1}{x+2} = \frac{1}{2+2} = \frac{1}{4}$

Answer: $\frac{1}{4}$

b) $\lim_{x \to 0} x \ln x$

Step 1: Rewrite using substitution

Let $x = e^{-t}$, so as $x \to 0^+$, $t \to \infty$. Then: $x \ln x = e^{-t} \ln(e^{-t}) = e^{-t}(-t) = -t e^{-t}$

Step 2: Evaluate the limit

We need: $\lim_{t \to \infty} -t e^{-t}$

Using L'Hôpital's Rule, rewrite $-t e^{-t}$ as: $\lim_{t \to \infty} \frac{-t}{e^t}$

Apply L'Hôpital's Rule (differentiate numerator and denominator): $\lim_{t \to \infty} \frac{-t}{e^t} = \lim_{t \to \infty} \frac{-1}{e^t} = 0$

Answer: $0$

c) $\lim_{x \to 0} \cos(2x)^{\frac{1}{x^2}}$

Step 1: Rewrite the expression

Let $y = \cos(2x)^{\frac{1}{x^2}}$. Take the natural logarithm: $\ln y = \frac{1}{x^2} \ln(\cos(2x))$

So: $\ln y = \lim_{x \to 0} \frac{\ln(\cos(2x))}{x^2}$

Step 2: Expand $\cos(2x)$ using a Taylor series

For small $x$, $\cos(2x) \approx 1 - 2x^2$. So: $\ln(\cos(2x)) \approx \ln(1 - 2x^2)$

Using $\ln(1 - u) \approx -u$ for small $u$: $\ln(\cos(2x)) \approx -2x^2$

Step 3: Simplify the limit

Substitute $\ln(\cos(2x)) \approx -2x^2$ into $\frac{\ln(\cos(2x))}{x^2}$: $\lim_{x \to 0} \frac{\ln(\cos(2x))}{x^2} = \lim_{x \to 0} \frac{-2x^2}{x^2} = -2$

So: $\ln y = -2 \implies y = e^{-2}$

Answer: $e^{-2}$

Summary of Answers:

a) $\frac{1}{4}$
b) $0$
c) $e^{-2}$

Would you like more detailed explanations for any of the steps?

Follow-up Questions:

1. How do Taylor series expansions help in evaluating limits?
2. Why is L'Hôpital's Rule valid for indeterminate forms?
3. Can we solve $\lim_{x \to 0} x \ln x$ using numerical approximations?
4. What happens if $\cos(2x)$ is replaced with $\sin(2x)$ in part (c)?
5. How does substitution simplify complicated limit problems?

Tip: For logarithmic and exponential limits, rewrite the function to isolate indeterminate forms and apply series expansions or L'Hôpital's Rule!