To solve the limit

$\lim_{x \to 1} \frac{7x^2 - 4x - 3}{3x^2 - 4x + 1},$

we first check by directly substituting $x = 1$ into both the numerator and the denominator:

Substituting $x = 1$ into the numerator:

$7(1)^2 - 4(1) - 3 = 7 - 4 - 3 = 0.$

Substituting $x = 1$ into the denominator:

$3(1)^2 - 4(1) + 1 = 3 - 4 + 1 = 0.$

Both the numerator and denominator are equal to 0, which means we have a $\frac{0}{0}$ indeterminate form. This suggests we can apply L'Hopital's Rule.

Step 1: Apply L'Hopital's Rule

L'Hopital's Rule states that if we have a $\frac{0}{0}$ indeterminate form, we can take the derivative of the numerator and the denominator separately, and then compute the limit of the resulting expression.

Derivative of the numerator:

The numerator is $7x^2 - 4x - 3$. Its derivative is:

$\frac{d}{dx}(7x^2 - 4x - 3) = 14x - 4.$

Derivative of the denominator:

The denominator is $3x^2 - 4x + 1$. Its derivative is:

$\frac{d}{dx}(3x^2 - 4x + 1) = 6x - 4.$

Step 2: Compute the new limit

Now, we compute the limit of the ratio of these derivatives:

$\lim_{x \to 1} \frac{14x - 4}{6x - 4}.$

Substitute $x = 1$:

$\frac{14(1) - 4}{6(1) - 4} = \frac{14 - 4}{6 - 4} = \frac{10}{2} = 5.$

Thus, the limit is:

$\boxed{5}.$

Follow-up Questions:

1. What is the general form of L'Hopital's Rule and when should it be applied?
2. How can we determine whether a limit leads to an indeterminate form before applying L'Hopital's Rule?
3. What happens if we continue applying L'Hopital's Rule multiple times in an indeterminate case?
4. How can we verify the limit using numerical approximations (e.g., plugging in values close to 1)?
5. What would happen if the numerator or denominator had higher-order terms (such as cubic terms)?

Tip: Always check for indeterminate forms first when evaluating limits; L'Hopital's Rule is a powerful tool but should be used after confirming that the form is indeterminate!