Evaluate the following limits : `lim_(x to infty)(sin(a/2^(x)))/sin(b/2^(x))`.

Answer

To evaluate the limit

\[ \lim_{x \to \infty} \frac{\sin\left(\frac{a}{2^x}\right)}{\sin\left(\frac{b}{2^x}\right)}, \]

we can follow these steps:

### Step 1: Analyze the limit as \( x \to \infty \)

As \( x \) approaches infinity, \( 2^x \) becomes very large. Therefore, \( \frac{a}{2^x} \) and \( \frac{b}{2^x} \) both approach 0.

### Step 2: Use the small angle approximation for sine

For small angles, we can use the approximation \( \sin(z) \approx z \) when \( z \) is close to 0. Thus, we can rewrite our limit:

\[ \sin\left(\frac{a}{2^x}\right) \approx \frac{a}{2^x} \quad \text{and} \quad \sin\left(\frac{b}{2^x}\right) \approx \frac{b}{2^x}. \]

### Step 3: Substitute the approximations into the limit

Substituting these approximations into our limit gives:

\[ \lim_{x \to \infty} \frac{\frac{a}{2^x}}{\frac{b}{2^x}} = \lim_{x \to \infty} \frac{a}{b} = \frac{a}{b}. \]

### Step 4: Conclusion

Thus, the limit evaluates to:

\[ \lim_{x \to \infty} \frac{\sin\left(\frac{a}{2^x}\right)}{\sin\left(\frac{b}{2^x}\right)} = \frac{a}{b}. \]