Quick DL Method Solution

Given:

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \]

LHS using derivative:

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \left.\frac{d}{dx}(x^4)\right|_{x=1} = 4x^3|_{x=1} = 4 \]

RHS using DL logic:

\[ \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \approx \frac{3k^2(x - k)}{2k(x - k)} = \frac{3k}{2} \]

Equating both sides:

\[ \frac{3k}{2} = 4 \Rightarrow k = \frac{8}{3} \]

\[ \boxed{k = \frac{8}{3}} \]

Evaluate: $$\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{1 - \cos x}$$

Step 1: Apply L'Hôpital's Rule (since it's 0/0):

First derivative: $$\frac{e^x + e^{-x} - 2}{\sin x}$$

Still 0/0 → Apply L'Hôpital's Rule again: $$\frac{e^x - e^{-x}}{\cos x}$$

Now, $$\lim_{x \to 0} \frac{1 - 1}{1} = 0$$

Final Answer: $$\boxed{0}$$