`underset(xrarr0)(lim)(sinnx)/x ` is equal to

Options

A. `n^2`
B. `1/n^2`
C. `n`
D. `1/n`

Answer

To solve the limit \(\lim_{x \to 0} \frac{\sin(nx)}{x}\), we can follow these steps:

### Step 1: Rewrite the limit
We start with the limit:
\[ \lim_{x \to 0} \frac{\sin(nx)}{x} \]

### Step 2: Multiply and divide by \(n\)
To make the expression easier to evaluate, we multiply and divide by \(n\):
\[ \lim_{x \to 0} \frac{\sin(nx)}{x} = \lim_{x \to 0} \frac{\sin(nx)}{nx} \cdot n \]

### Step 3: Use the standard limit
We know from calculus that:
\[ \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \]
In our case, let \(u = nx\). As \(x\) approaches \(0\), \(u\) also approaches \(0\). Therefore, we can rewrite our limit:
\[ \lim_{x \to 0} \frac{\sin(nx)}{nx} = 1 \]

### Step 4: Substitute back into the limit
Now substituting this result back into our expression, we have:
\[ \lim_{x \to 0} \frac{\sin(nx)}{x} = n \cdot 1 = n \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\sin(nx)}{x} = n \]

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