Q.Evaluate $\lim_{x \to a} \dfrac{\sin 3x}{\sin 7x}$.

The key idea here is the continuity of trigonometric functions.

1. The functions $\sin 3x$ and $\sin 7x$ are continuous for all real values of $x$.
2. The given function, $f(x) = \frac{\sin 3x}{\sin 7x}$, is a ratio of two continuous functions.
3. A ratio of continuous functions is continuous at any point $x=a$ where the denominator is non-zero. …

When evaluating limits of functions, especially those involving quotients, the first step is always to understand the nature of the function and the point to which $x$ is approaching. Our function $f(x) = \frac{\sin 3x}{\sin 7x}$ is a quotient of two continuous functions, $\sin 3x$ and $\sin 7x$.

Concept: Limits of Continuous Functions

For a function $f(x)$ that is continuous at a point $a$, the limit as $x$ approaches $a$ is simply the function's value at $a$, i.e., $\lim_{x \to a} f(x) = f(a)$.

When dealing with a quotient of two continuous functions, say $f(x) = \frac{g(x)}{h(x)}$, the limit can be found by direct substitution, $\frac{g(a)}{h(a)}$, provided that the denominator $h(a)$ is not zero.

If $h(a) = 0$, we encounter one of two situations:

1. $\frac{\text{non-zero}}{0}$ form: This indicates that the limit does not exist, and the function typically approaches $\pm \infty$.
2. $\frac{0}{0}$ indeterminate form: This means direct substitution fails, and we need to use other techniques like L'Hopital's Rule, series expansion, or standard limit formulas to evaluate the limit.

Let's apply this understanding to the given problem.

Step-by-Step Evaluation

1. Initial Check: Direct Substitution We attempt to substitute $x=a$ into the expression:

Now, we must consider the value of the denominator, $\sin 7a$.

2. Case 1: The Denominator is Non-Zero ($\sin 7a \neq 0$)

If $\sin 7a \neq 0$, then the denominator is not zero, and the function is continuous at $x=a$. In this scenario, direct substitution is valid and gives us the limit directly.

> [!IMPORTANT]

> If $\sin 7a \neq 0$, then $\lim_{x \to a} \frac{\sin 3x}{\sin 7x} = \frac{\sin 3a}{\sin 7a}$.

3. Case 2: The Denominator is Zero ($\sin 7a = 0$)

   If $\sin 7a = 0$, this means $7a$ must be an integer multiple of $\pi$. So, $7a = n\pi$ for some integer $n \in \mathbb{Z}$. This implies $a = \frac{n\pi}{7}$.

   In this case, direct substitution leads to a problematic form. We need to further analyze the numerator, $\sin 3a$.

   • Subcase 2a: Numerator is Non-Zero ($\sin 3a \neq 0$)

     If $\sin 7a = 0$ but $\sin 3a \neq 0$, then the limit takes the form $\frac{\text{non-zero}}{0}$.

     For example, if $a = \frac{\pi}{7}$, then $\sin 7a = \sin(\pi) = 0$. However, $\sin 3a = \sin(\frac{3\pi}{7}) \neq 0$.

     In such situations, the limit does not exist, as the function's magnitude approaches infinity.

     Watch out

     If $\sin 7a = 0$ but $\sin 3a \neq 0$, the limit $\lim_{x \to a} \frac{\sin 3x}{\sin 7x}$ does not exist.

   • Subcase 2b: Numerator is Zero ($\sin 3a = 0$)

     If both $\sin 7a = 0$ and $\sin 3a = 0$, then the limit takes the indeterminate form $\frac{0}{0}$.

     For $\sin 3a = 0$, we must have $3a = m\pi$ for some integer $m \in \mathbb{Z}$.

     So, we have two conditions: $7a = n\pi$ and $3a = m\pi$.

     This implies $a = \frac{n\pi}{7}$ and $a = \frac{m\pi}{3}$.

     Equating these, $\frac{n\pi}{7} = \frac{m\pi}{3} \implies \frac{n}{7} = \frac{m}{3} \implies 3n = 7m$.

     Since 3 and 7 are coprime, $n$ must be a multiple of 7, and $m$ must be a multiple of 3.

     Let $n = 7k$ and $m = 3k$ for some integer $k \in \mathbb{Z}$.

     Substituting back into $a = \frac{n\pi}{7}$, we get $a = \frac{7k\pi}{7} = k\pi$.

     Similarly, from $a = \frac{m\pi}{3}$, we get $a = \frac{3k\pi}{3} = k\pi$.

     Thus, the $\frac{0}{0}$ indeterminate form occurs precisely when $a = k\pi$ for any integer $k$. This includes the common case where $a=0$ (when $k=0$).

     To evaluate the limit in this $\frac{0}{0}$ case ($a=k\pi$):

     • Method A: Using Standard Limit Formula (for $a=0$)

       This method is particularly useful when $a=0$.

       The fundamental trigonometric limit is $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

       A useful extension is $\lim_{x \to 0} \frac{\sin kx}{kx} = 1$ for any non-zero constant $k$.

       If $a=0$, the limit is $\lim_{x \to 0} \frac{\sin 3x}{\sin 7x}$.

       We can manipulate the expression to use the standard limit formula:

       $\lim_{x \to 0} \frac{\sin 3x}{\sin 7x} = \lim_{x \to 0} \left( \frac{\sin 3x}{3x} \cdot \frac{7x}{\sin 7x} \cdot \frac{3x}{7x} \right)$ …