It seems the screenshot is attempting to solve the limit $\lim_{x \to 0} \frac{\sin(7x)}{\sin(6x)}$, and the direct substitution approach is producing an indeterminate form.

The Correct Approach:

1. Substitute $x = 0$ directly into the limit:

   • $\sin(7x) \to 0$ and $\sin(6x) \to 0$, hence $\frac{0}{0}$, which is an indeterminate form.

2. Apply L'Hôpital's Rule: Since we have an indeterminate form of $\frac{0}{0}$, we can differentiate the numerator and denominator.

   • The derivative of $\sin(7x)$ is $7 \cos(7x)$,
   • The derivative of $\sin(6x)$ is $6 \cos(6x)$.

   Now the limit becomes: $\lim_{x \to 0} \frac{7 \cos(7x)}{6 \cos(6x)}.$

   At $x = 0$:

   • $\cos(7(0)) = 1$,
   • $\cos(6(0)) = 1$.

   Thus, the limit is: $\frac{7 \cdot 1}{6 \cdot 1} = \frac{7}{6}.$

Explanation of the Error:

The error occurs because, instead of directly substituting the constants $7$ and $6$ from the sine functions, the limit needed the application of a rule (L'Hôpital's or simplification using small-angle approximations) to handle the indeterminate form.

Would you like further details or clarifications on this process?

Here are 5 related questions to expand your understanding:

1. How does L'Hôpital's Rule work, and when can it be applied?
2. What is the small-angle approximation for sine and how could it be applied here?
3. Can this limit be solved using series expansion of sine functions?
4. What are other indeterminate forms besides $\frac{0}{0}$ and how are they handled?
5. Can limits like this one be solved graphically to verify the result?

Tip: When encountering indeterminate forms like $\frac{0}{0}$, consider using L'Hôpital's Rule or approximations like Taylor series to simplify the limit.