Let's analyze the problem provided in the images. We are tasked with evaluating the following limit:

$\lim_{x \to 1} \frac{x^{1505} - 1}{x - 1}$

Part (a)

To solve this limit, we notice that directly substituting $x = 1$ results in the indeterminate form $\frac{0}{0}$. So, we need to use algebraic or calculus techniques to resolve this limit.

One method is to use L'Hopital's Rule, which applies when we have an indeterminate form like $\frac{0}{0}$. The rule states that:

$\lim_{x \to 1} \frac{f(x)}{g(x)} = \lim_{x \to 1} \frac{f'(x)}{g'(x)}$

if the limit on the right exists. Here:

• $f(x) = x^{1505} - 1$
• $g(x) = x - 1$

We calculate the derivatives:

• $f'(x) = 1505x^{1504}$
• $g'(x) = 1$

Now applying L'Hopital's Rule:

$\lim_{x \to 1} \frac{1505x^{1504}}{1} = 1505 \cdot 1^{1504} = 1505$

So, the value of the limit is:

$\lim_{x \to 1} \frac{x^{1505} - 1}{x - 1} = 1505$

Part (b)

In part (b), the substitution $h = x - 1$ is suggested. Substituting this into the limit:

$x = h + 1$

The limit becomes:

$\lim_{h \to 0} \frac{(h + 1)^{1505} - 1}{h}$

This is equivalent to the original limit, but now in terms of $h$. As $h \to 0$, this limit simplifies similarly to the previous step, leading again to the same result:

$\lim_{h \to 0} \frac{(h + 1)^{1505} - 1}{h} = 1505$

Thus, both methods lead to the same final value of the limit, which is 1505.

Would you like further details on any part of this solution?

Here are five related questions for further exploration:

1. How is L'Hopital's Rule applied to more complex indeterminate forms?
2. Can this problem be solved using the binomial expansion instead of L'Hopital's Rule?
3. What is the significance of the derivative in finding limits like this?
4. How does the substitution $h = x - 1$ affect the structure of the limit?
5. What other calculus techniques are useful for resolving indeterminate forms?

Tip: When using L'Hopital's Rule, make sure the limit results in an indeterminate form like $0/0$ or $\infty/\infty$ before applying the rule.