Let's solve both limits shown in the image.

Part (a):

$\lim_{x \to 1} \frac{x^4 - 1}{x - 1}$ First, notice that if you substitute $x = 1$, you get the indeterminate form $\frac{0}{0}$. To resolve this, we can factor the numerator using the difference of squares:

$x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1)$

Thus, the limit becomes: $\lim_{x \to 1} \frac{(x - 1)(x + 1)(x^2 + 1)}{x - 1}$

We can cancel out the $(x - 1)$ terms, leaving us with: $\lim_{x \to 1} (x + 1)(x^2 + 1)$

Now, substitute $x = 1$ into the remaining expression: $(1 + 1)(1^2 + 1) = 2 \times 2 = 4$

So, the limit for part (a) is: $\boxed{4}$

Part (b):

$\lim_{x \to 0} \frac{e^{2x} - 1}{x}$ Again, if you substitute $x = 0$, you get the indeterminate form $\frac{0}{0}$, so we can apply L'Hopital's Rule. L'Hopital's Rule states that if you have an indeterminate form, you can differentiate the numerator and the denominator separately and then take the limit.

1. Differentiate the numerator: $\frac{d}{dx}(e^{2x} - 1) = 2e^{2x}$
2. Differentiate the denominator: $\frac{d}{dx}(x) = 1$

Now, the limit becomes: $\lim_{x \to 0} \frac{2e^{2x}}{1} = 2e^0 = 2$

So, the limit for part (b) is: $\boxed{2}$

Would you like any further details or have any questions? Here are 5 related questions you might explore:

1. How does factoring help solve limits involving polynomials?
2. Can L'Hopital's Rule always be applied to indeterminate forms like $\frac{0}{0}$?
3. What is the difference between applying L'Hopital's Rule and using a Taylor expansion to solve limits?
4. How do we handle more complicated expressions that lead to indeterminate forms?
5. Can you think of a scenario where a limit does not exist?

Tip: Always check if an expression results in an indeterminate form before applying L'Hopital’s Rule.