https://file.aiquickdraw.com/mathbot/file/17260000094459akq1qe5.jpg

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.

Calculate the following limits: 
(a) \( \lim_{x \to 1} \frac{x^4 - 1}{x - 1} \) 
(b) \( \lim_{x \to 0} \frac{e^{2x} - 1}{x} \)

Learn how to solve limits involving indeterminate forms such as 0/0 using factoring and L'Hopital's Rule. This tutorial covers examples like \( \lim_{x \to 1} \frac{x^4 - 1}{x - 1} \) and \( \lim_{x \to 0} \frac{e^{2x} - 1}{x} \). Suitable for students studying advanced high school or college calculus.

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