The value of lim (n sin 5x) / (sin 3x) as x approaches 0 is
Understand the Problem
The question is asking for the value of a limit as x approaches 0 for the expression (n sin(5x)) / (sin(3x)). This requires applying limit properties and potentially L'Hôpital's rule if the limit evaluates to an indeterminate form.
Answer
The answer is $\frac{5n}{3}$.
Answer for screen readers
The final answer is: $$ \frac{5n}{3} $$
Steps to Solve
- Set up the limit expression
We want to evaluate the limit: $$ \lim_{x \to 0} \frac{n \sin(5x)}{\sin(3x)} $$
- Apply L'Hôpital's Rule
Since substituting $x = 0$ leads to the form $\frac{0}{0}$, we can apply L'Hôpital's Rule. This requires differentiating the numerator and denominator:
- The derivative of the numerator $n \sin(5x)$ is $n \cdot 5 \cos(5x)$.
- The derivative of the denominator $\sin(3x)$ is $3 \cos(3x)$.
So, we will evaluate the limit: $$ \lim_{x \to 0} \frac{n \cdot 5 \cos(5x)}{3 \cos(3x)} $$
- Evaluate the limit as x approaches 0
Substituting $x = 0$ into the differentiated expression: $$ \frac{n \cdot 5 \cos(0)}{3 \cos(0)} = \frac{n \cdot 5}{3} $$
Since $\cos(0) = 1$.
- Write the final result
Thus, we find: $$ \lim_{x \to 0} \frac{n \sin(5x)}{\sin(3x)} = \frac{5n}{3} $$
The final answer is: $$ \frac{5n}{3} $$
More Information
This limit utilizes the relationship $\lim_{u \to 0} \frac{\sin(u)}{u} = 1$, which helps simplify trigonometric expressions as they approach zero.
Tips
- Not recognizing the indeterminate form $\frac{0}{0}$ and skipping L'Hôpital's Rule.
- Forgetting to differentiate both the numerator and denominator when applying L'Hôpital's Rule.
