Let's apply L'Hôpital's Rule to evaluate the limit.

Given limit:

$\lim_{{x \to 2}} \frac{x + 47 - 7}{x - 2}$

First, simplify the expression:

$\lim_{{x \to 2}} \frac{x + 40}{x - 2}$

As we discussed earlier, directly substituting $x = 2$ gives a division by zero, indicating that L'Hôpital's Rule may be applied. L'Hôpital's Rule states that if a limit results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, we can evaluate the limit of the derivatives of the numerator and denominator.

Differentiate the numerator and the denominator separately with respect to $x$:

1. The derivative of the numerator $x + 40$ is $1$.
2. The derivative of the denominator $x - 2$ is also $1$.

Now, apply L'Hôpital's Rule:

$\lim_{{x \to 2}} \frac{d/dx \, (x + 40)}{d/dx \, (x - 2)} = \lim_{{x \to 2}} \frac{1}{1} = 1$

Thus, the limit is:

$\lim_{{x \to 2}} \frac{x + 40}{x - 2} = 1$

Would you like more details or have any questions?

Here are 5 related questions you might find helpful:

1. When should L'Hôpital's Rule be applied?
2. Can L'Hôpital's Rule be applied to limits involving $\infty$?
3. What are other methods to evaluate limits besides L'Hôpital's Rule?
4. How does L'Hôpital's Rule apply to higher-order derivatives?
5. What are common pitfalls when using L'Hôpital's Rule?

Tip: Always check that the limit is in an indeterminate form before applying L'Hôpital's Rule. If it is not, L'Hôpital's Rule cannot be applied correctly.